Mathematica, 19 bytes

Median[#~Riffle~0]&

Code and golfing thanks to Martin Büttner.

This is an unnamed pure function that takes in a list of integers as input. Invoke it like

Median[#~Riffle~0]&[{-2, -3, -2, -4}]

or similarly saved to variable.

The code first riffles a zero in between every two elements of the input list, which inserts n-1 zeroes among n elements. Then, it takes the median to produce the answer.

This gives min-mod because it handles each case:

1. All the numbers are positive, in which case the zeroes are below them and the median is the lowest positive number.

2. All the numbers are negative, in which case the zeroes are above them and the median is the least negative number.

3. There's both a positive and negative number, and so the middle element is a zero.

If Mathematica implements its median using the linear-time selection algorithm, then this is also O(n).

Haskell, 62 61 39 38 bytes

f s=last$0:[x|x<-s,and[x*x<=x*y|y<-s]]

using some comparison magic borrowed from @Zgarb's answer* , namely, x*x<=x*y.

x*x<=x*y is true only when x and y have the same sign and y's absolute value is bigger. note that when x is 0 it is always true.

we determine that x is the result iff it is contained in s, and that for all y in s x has the same sign as y and is smaller in absolute value. if no value in s satisfies this definition, then 0 is the result.

f then works by searching s for an element to satisfy this, and uses 0 as a default.

_{*though he didn't use it for the reasons I'm using it, and he actually got rid of it by now}

JavaScript (ES6), 39 bytes

a=>a.reduce((p,c)=>p*c>0?p*p>c*c?c:p:0)

GolfScript, 10 9 bytes

~{0]$1=}*

Assumes input from stdin in the format [-4 -2 -3 -2]

This uses the built-in sort function $, but every time it's invoked it's on an array of 3 elements, which is permitted.

Online demo

Python 2, 53

lambda l:reduce(lambda a,b:sorted([a,b,0])[1],l,l[0])

The idea is to use reduce to turn the two-input min-mod finder into an n-input one. I came up with it independently of the other answers that use it. Only Python 2 supports reduce.

The two-input solution simply finds the median of the two numbers and zero. See my Mathematica answer for a more direct way to use the median.

Less golfed:

def f(l):
 A=l[0]
 for x in l:A=sorted([a,b,0])[1]
 return A

A hypothetical amalgam of Python 2 and Python 3 would be a character shorter, with the starred assignment from Python 3 and input() and print from Python 2.

#Not real code!
A,*l=input()
for x in l:A=sorted([A,x,0])[1]
print A

Old code, without sorting:

lambda l:reduce(lambda a,b:[a,b][a*a>b*b]*(a*b>0),l,l[0])

Haskell, 83 40 39 bytes

This is probably not the shortest possible Haskell solution (and certainly won't beat the others here), but it's a start. EDIT: Now over 50 % shorter! EDIT2: One byte less...

a#b|a*b<0=0|a*a<b*b=a|1<2=b
m=foldr1(#)

This is just a straightforward fold (or reduce, as some languages call it) by the binary operator #, which computes the median of a, b and 0. Even though the rules would now allow me to sort small lists, this requires an import in Haskell and results in a higher byte count (49 bytes, but 31 without the import):

import Data.List
a#b=sort[a,b,0]!!1
m=foldr1(#)

CJam, 20 bytes (or 10 bytes)

q~{]__~z\z<=\~*0>*}*

Using @xnor's approach, reduce calculating minmod of 2 numbers at a time from the array.

This would have been 19 bytes if :z worked

Using the new rule of using sorts on short arrays:

q~{0]$1=}*

which is exactly equivalent to @Peter's answer

Previous 26 bytes asnwer:

q~_{g}%_|:+\(z\{za+_~>=}/*

This can be golfed further...

Input (via STDIN) is the integer array like :

[-4 -2 -3 -2]

and output is the minmod of the input array

Try it here

If only :g and :z worked, this would have been 4 bytes shorter.

TI-BASIC, 12 bytes

Assumes input in format of {-2,4,3}

dim(Ans)2-1→dim(Ans:median(Ans

List of token sizes here, hex dump below:

B5 72 11 32 B0 31 04 B5 72 3E 1F 72

Works similarly to xnor's answer:

dim(Ans)    get the length of the array
2-1         multiply by 2 and subtract 1
→dim(Ans    make this the new length (new elements always default to 0)
:           new command (since it is last one it will return the value)
median(Ans  calculates and returns the median of the new array

Marbelous, 210 bytes

@0
00
]]\\&002
/\..//&0@0
00..02
MMMMMM//\\
:M
}0}1}0}1}0}1}0}2..}2
^7^7||||&0&1&4<3&0=2{>
EqalLteq{0{<{<<2&1--
&2..&3..}100..&2\/{>
>0&6=0&4&5&6..\/
&3..&5\/{<{0
\/..\/
:|
}000}0
&0Subt
{0&1
}0{0
^7
=0&1
&0
\/

There are three boards used here.

The | board (Ab in the readable version) takes the absolute value of a marble (by either returning the passed marble or zero minus the passed marble, as all arithmetic in Marbelous is unsigned).

The M board (Minabs in the readable version) finds and outputs to the left either the first or second marble passed (whichever has a smaller absolute value), and exiting if a different signed marble is passed.

The M board also releases the marble it holds downward instead of leftward once the last character from STDIN is fetched.

The M board is used in the main board to store the minmod of all checked values at any given time, as it releases the value to be saved leftward, which is then deflected back in.

Trashbins (\/) were only placed under synchronisers that would otherwise print to STDIN.

Input/Output uses STDIN/STDOUT. Both deal with 8-bit values (if you wanted to pass +0x30 and +0x38, place 08 into STDIN).

Libraries and cylindrical boards are both required. Viewing output as decimal numbers is recommended (note that this displays the unsigned value of the minmod result).

Test it here.

Note: For more human-friendly input/output, add Dp under the last line of the main board (before :M), replace ]] with Rd, and add the following at the bottom:

:Rd
}0}0}0
]]]]]]{>
-O-O-O
-O-O-O
*A
Plus
\\*A
..Plus
..{0
:*A
}0}0
<<<<
<<
<<
Plus
{0

This simply changes the output to be 3 decimal digits. Likewise, input with these changes requires a space separated list of 3 decimal digits per number.

Readable Version:

Python 2, 82 79 71 69 61 bytes

lambda l:reduce(lambda G,H:[H,G][(G>H)^(G>0)]*(G*H>0),l,l[0])

This is based off of my pyth answer, which was inspired by Mig's answer.

Old answer:

l=input()
m=l[0]
k=1-2*(m<0)
for i in l:m=[m,i][m>i*k]
print(k*m>0)*m

This is a very long answer. I feel like having 2 variables is a waste...? I was right...? ish? ;p

KDB/Q, 43 characters for function body definition

Thanks to great ideas from previous posts:

f:{$[all 1_0<(*':)x;{$[<[x*x;y*y];x;y]}/[x];0]}

Enter single number using enlist

f[enlist 2]
f[enlist 0]
f[enlist -2]
f[2 4 1]
f[0 1 2]
f[1 0 2]
f[-1 1 2]
f[-4 -2 -3 -2]
f[-5 0 -1]
f[-5 -0 -1]
f[1 0 -1]

I'm sure some Q guru can come up with shorter ones.

Pyth, 25 22 20 12

uhtS[0GH)QhQ

Probably not novel, but original :P

Pre-sorting allowed

u*?Gx>GH>G0H>*GHZQhQ

Pyth

Try it online.

The idea to use reduce and ternary statements was shamelessly stolen from Mig's answer, but I have no idea if these algorithms are otherwise even similar, as I can't read ternary statements.

Explanation:

Q=eval(input)         : implicit
u                QhQ  : print reduce(lambda G,H: ..., Q, Q[0])
 *          >*GHZ     : ... * (G*H>0)
  ?G       H          : G if ... else H
    x>GH>G0           : G>H xor G>0

C#, 101 bytes

My first try at code golfing and in a pretty golfing hostile language. Based on reduce (Aggregate in LINQ) and very similar to the JavaScript answer by Mig. Can be run like (new System.Linq.M()).m(new[] {1, 2, 3}). Passes all test cases, but doesn't handle empty input arrays.

namespace System.Linq{class M{public int m(int[]i){return i.Aggregate((a,b)=>a*b>0?a*a<b*b?a:b:0);}}}

Game Maker Language, 489 bytes

About Game Maker Language

Riffles the array (zeros are appended) and returns the median (similar to my other answer)

i=0a=argument0
while(variable_local_array_get(a,i))i++
for(j=0;j++;j<i-1)a[j+i]=0var i,j,d,m=0d=ds_list_create()if variable_local_exists(a){if variable_local_array_get(a,0){for(i=0;i<32000;i++){if variable_local_array_get(a,i)=0break
ds_list_add(d,variable_local_array_get(a,i))}ds_list_sort(d,0)i=ds_list_find_value(d,ds_list_size(d) div 2)j=ds_list_find_value(d,(ds_list_size(d) div 2)-1)m=ds_list_find_value(ds,ds_list_size(d) mod 2)ds_list_destroy(d)}if m return (i+j)/2return i
break}

Java, 84 bytes

This is Java in all its glory. Beats GolfScript by a factor of slightly over 900%.

int f(int[]a){int b=a[0],c;for(int d:a)b=(c=d<0?-1:1)*b<0?0:d*c<b*c?d:b;return b;}

Wrapped in class:

public class MinModGolfed{

    public static void main(String[] args){
        int[] numbers = new int[args.length];
        for (int i = 0; i < args.length; i++){
            numbers[i] = Integer.parseInt(args[i]);
        }
        System.out.println(new MinModGolfed().f(numbers));
    }

    int f(int[]a){int b=a[0],c;for(int d:a)b=(c=d<0?-1:1)*b<0?0:d*c<b*c?d:b;return b;}

}

Expanded with comments:

public class MinModExpandedGolfed{

    public static void main(String[] args){
        int[] numbers = new int[args.length];
        for (int i = 0; i < args.length; i++){
            numbers[i] = Integer.parseInt(args[i]);
        }
        System.out.println(new MinModExpandedGolfed().f(numbers));
    }

    int f(int[]a){                  //a is the input numbers
        int b=a[0],c;             //b is the best number found so far.
        for(int d:a)               //Iterate over a with current element as d.
            b=(c=d<0?-1:1)         //c is equal to the sign of d.
                    *b<0?
                        0:          //If b has opposite sign of d, b = 0.
                        d*c<b*c?d:b;//If the absolute value of d is less than b, b = d. 
        return b;
    }

}

Note: This can be improved using Java 8.

Note: Effort to improve in Java 8 failed.

GolfScript, 18 bytes

~](.@{]$~\;}/0]$1=

Test this code online.

This program reads the input from stdin as a whitespace-separated list of integers, and prints the minmod of these integers to stdout. It passes all the test cases given above, but does require the input list to be non-empty.

Explanation:

This program is based on the following identity:

        minmod ⟨x_i⟩ = median( 0, min ⟨x_i⟩, max ⟨x_i⟩ )

where ⟨x_i⟩ = (x₁, x₂, ..., x_n) denotes the input list.

Since we're only allowed to sort fixed-length lists, this code finds the minimum and maximum of the list by repeatedly sorting a three-element list, throwing out the middle element and replacing it with the next input value. At the end, the middle element is replaced by 0, the three-element list is sorted once more, and the middle value is returned.

Matlab/Octave, 26

This is basically just a translation of the Mathematica answer by xnor. It works by appending one zeros less than the length of the input vector. Note that appending one more would not work, since then the result would be 0 all the time. Thanks to MartinBüttner for the -4 chars of this solution=)

@(x)median([x,0*x(2:end)])

Java, 353 304 124 bytes

Put together the worst language for code golf with the world's worst golfer and you get...

int m(int[]a){int m=a[0];if(m<0)for(int i:a){m=(i>m)?i:m;m=(i>0)?0:m;}else for(int i:a){m=(i<m)?i:m;m=(i<0)?0:m;}return m;}}

Ungolf it and you get:

int m(int[] a) {
    int m = a[0];
    if (m < 0) {
        for (int i : a) {
            m = (i > m) ? i : m;
            m = (i > 0) ? 0 : m;
        }
    } else {
        for (int i : a) {
            m = (i < m) ? i : m;
            m = (i < 0) ? 0 : m;
        }
    }
    return m;
}

This is a function (if it wasn't pretty damn obvious) that receives an array of numbers and processes its values, returning the minmod value.

My old benemoth of a sollution is also included, which is a whole program - as always.

class M{public static void main(String[]a){java.util.Scanner s=new java.util.Scanner(System.in);int n,m=0;try{m=s.nextInt();if(m<0)while(true){n=s.nextInt();m=(n>m)?n:m;m=(n>0)?0:m;}else while(true){n=s.nextInt();m=(n<m)?n:m;m=(n<0)?0:m;}}catch(java.util.InputMismatchException e){System.out.print(m);}}}

Ungolf it and you get:

class M {

    public static void main(String[] a) {
        java.util.Scanner s = new java.util.Scanner(System.in);
        int n = 0, m = 0;
        try {
            m = s.nextInt();
            if (m < 0) {
                do {
                    n = s.nextInt();
                    m = (n > m) ? n : m;
                    m = (n > 0) ? 0 : m;
                } while (true);
            } else {
                do {
                    n = s.nextInt();
                    m = (n < m) ? n : m;
                    m = (n < 0) ? 0 : m;
                } while (true);
            }
        } catch (java.util.InputMismatchException e) {
            System.out.print(m);
        }
    }
}

Receives infinite numbers, stops when a non-number value is entered, presenting the Minmon value.

J, 20 12 bytes

Function taking the list as argument. Stolen from the Golfscript/CJam/whatever.

(1{0/:~@,,)/

The minmod of x and y is the median (sort /:~ and take the middle 1{) of the three item list 0,x,y. Reduce the list (folding in J parlance) by taking this minmod between adjacent elements.

In use at the REPL. (J spells its negative sign _.)

   (1{0/:~@,,)/ _4 _2 _3 _2
_2
   f =: (1{0/:~@,,)/    NB. give it a name
   f 1 1 2
1
   f 0 1 2
0
   f _1 1 2
0

Old garbage, before I noticed short sorts are allowed: 0:`<.`>.@.(*@]*0<*))/ The minmod of x and y is 0 (0:) if 0 is greater than or equal to the product of x and y, else it is the min (<.) or the max (>.) between x and y depending on the sign. Fold this over the whole list.