I don't know the solution to this problem, but here are a few things that might be useful:
There are only 2m distinct absolute values available between -(2m-1) and +(2m-1). Therefore if you have 2m+1 distinct numbers, then an absolute value must be repeated. This means that there must be a positive and negative pair. In other words, the set must contain two distinct values j and -j. If the set also contains zero, then you're done.
Therefore, we may assume that zero is not one of the 2m+1 distinct integers. This means that there are in fact only 2m-1 possible absolute values (since zero is excluded). Therefore there must be two repeated absolute values, i.e. two positive/negative pairs.
I haven't worked out where to go from here, and it might be a dead end. It might be useful to play around with a small concrete case, say m=4 or 5, and see how you might prove it in that case.
