Actually, if I have this figured out correctly, 26 integers are enough.
Let’s say the 26 integers are . We may assume without loss of generality that if x is in the set then –x is not; otherwise we would have x + (-x) = 0, which is divisible by 50, and we would be through. Consider the expanded set made by throwing in the negatives of the original integers, i.e.
This set contains at least 2 * 26 – 1 = 51 distinct integers, taking into account the possibility that one of the integers might be 0, in which case x and –x are the same.
Now place the integers into pigeonholes based on their residue class mod 50, i.e. if the remainder on dividing x by 50 is j, then place x in the jth pigeonhole. Then there are 50 pigeonholes, numbered 0 to 49, so some pigeonhole must contain at least two integers, say and . If the two ’s are the same (two +’s or two –‘s), then is divisible by 50. If the two ’s are different (+ and – or – and +), then is divisible by 50.