In any set of 181 square integers, prove that one can always find a
subset of 19 numbers, the sum of whose elements is divisible by 19.
I'll leave it to someone a little more mathmatically rigid to refine my thoughts if they wish.
The fact that the numbers are perfect squares is extra information. Any group of number greater than "n", must have at least one subset of "n" members or less whose sum is divisible by n. This is because there are n-1 possible remainders when dividing the terms by n.
Of course, to ensure that there is a combination with "Exactly" 19 terms (in this case), requires that some duplicate remainders are present, to cancel each other out once the multiple of 19 is reached.
conceptual solution
