I first solve a related but easier puzzle. My related puzzle is, you start with the 2^n vectors with entries +1 or -1. Child changes some subset of entries to 0s. Then I will sketch proof that the new set of 2^n vectors always contains either two disjoint subsets with equal sums, or two with sums that are negatives of each other. (Asimov had wanted only the latter.) It's easy once you see this trick. You just consider the number of possible ways to select two disjoint subsets, which is doubly-exponentially huge, namely 3^(2^n). Then you consider the possible vector sums they could have. This number of possibilities is way smaller, namely at most 2^((n+1)*n). This makes it obvious if n is large enough then necessarily two subsets must exist with equal or negated vector-sums (regardless of what that child had done). On the other hand, if n is too small for this proof to work, then you have a small problem you can solve by brute force (computer?). I'd pretty much already shown Asimov was right when n<=3 by hand (although I admit "pretty much" is not a real proof) which would suffice. QED. OK, now, what about Asimov's original harder form of the puzzle? Perhaps it can be solved with the same kind of trick, but you'll need to add some new ingredient also.