1) There are 90 numbers, so we have 2^90 subsets.
2) All numbers are < 10^25, so the sum of all 90 numbers is < 90 * 10^25 < 10^27. So each sum of a subset is an integer in the range of 0 to 10^27-1. So we have ≤ 10^27 possible sum values.
Since 2^90 > 10^27, we have more subsets than possible sum values. Hence, there are at least two subsets having the same sum.
BTW, this is a beautiful application of the Pigeonhole principle: https://en.wikipedia.org/wiki/Pigeonhole_principle
Edit: wait I think I've got it:
Each subset implicitly also chooses it's counter subset, e.g. if you choose the subset consisting of the first 45 numbers, you've also therefore said the other subset consists of the last 45 numbers. Since there are more of these "dual subsets" than possible sums, the pigeonhole principle yadda-yadda.
You are not forced to use all numbers. (And in fact, my explicit solution doesn't use all numbers.) There is no restriction that the union of both subsets must be equal to the full set. Note that this restriction would change the question dramatically. See also: https://news.ycombinator.com/item?id=10023098
Also note that it is allowed for both chosen subsets to overlap. In fact, the proof just says that two different (i.e. not entirely equal) subsets with the same sum exist.
However, once you have a pair of different overlapping subsets of the same sum, you can simply remove the intersection from both sets. Both sums decrease by the same amount. You then get a pair of two disjoint sets that have the same sum.
And no, I don't think the proof would not work when requiring the two subsets to form a partition. Also note that for partitions, the number of sums would be one, as the only possible sum is half of the sum of all numbers. (And the problem would be unsolvable if the sum of all numbers was odd, but all numbers were still integers.)
(I didn't realize you were allowed to use each number in both subsets, so I have no idea if I would have figured it out or not on my own.)
Why? (Obviously you wouldn't compute it so what's the "trick"?)