I said:
We'll be done if we can find u(n) when n is a product of distinct primes. I don't see how to do that right now, but I suspect it's not all that hard. (Famous last words.)
Not-yet-proven observation: if p is an odd prime then u(2p) = 2^p+2. Explanation: The zero-sum configurations are all the antipodal ones (of which there are 2^p) and two regular p-gons (of which there are 2). Generalization: Let n = p1...pk, a product of distinct primes. Let A(j,k) consist of vertices k, k+n/pj, k+2n/pj, ..., k+(pj-1)n/pj. Clearly A(j,k) has sum 0. So does the union of any set of disjoint A(j,k). Conjecture: These are *all* the zero-sum configurations. That A(j,k) has sum 0 is trivial. So is the observation that the set of zero-sum sets is closed under disjoint unions. So we need "only" show that every zero-sum configuration arises in this way. Let A be any zero-sum configuration. If A contains any A(j,k) then A \ A(j,k) still has sum 0, so by induction on the size of A we may assume that A *doesn't* contain any A(j,k). Is there some obvious reason why in fact A must be empty? I don't see one. But let's return to the special case n=2m, m odd and squarefree. (We may need to restrict further to prime m; let's see.) Then the standard 2m-gon is composed of the standard m-gon and its negation. So suppose we have a 0-sum subset of that that isn't antipodal; that means we have two different subsets of the standard m-gon with equal sum. One way to do this is to find two with sum 0, and that's where the "regular p-gon" configurations mentioned above for n=2p come from. Is that all? It is when m=p, because otherwise we get two different monic polynomials of degree at most p-1 of which w(p) is a root, which is impossible since the degree of w(p) is p-1. That proves the observation above: u(2p) = 2^p+2. I have to go and do other things now. Anyone want to finish this off? :-) -- g