From: "Meeussen Wouter (bkarnd)" <wouter.meeussen@vandemoortele.com>
it is a small surprise that, for w = Exp[ 2 Pi I /30 ] alias 1^(1/30) , w+w^7+w^8+w^18+w^19+w^25 = 0 no cancellation of symmetric pairs, triples or quintuples in sight, but w+w^7+w^19+w^25 = -w^13 (* four out of five *) and w^8+w^18 = -w^28 = +w^13 (* two out of three *)
could be called "cancellation by missing parts" ;-))
Let #(N) be the number of zero sum subsets of complex N'th roots of unity. I calculate #(30) = 146854 and more generally #(2.3.p) = 10^p + 6 (6^p + 2^p + 1) for prime p not 2 or 3, and #(2.5.p) = 34^p + 10 (18^p + 2.12^p + 2.8^p + 6.4^p + 7.2^p + 3) for prime p not 2 or 5. More precisely, I find at least that many zero sum subsets. I conjecture that's all the zero sum subsets, but I don't have a proof. When N contains 3 or more distinct prime factors, "cancellation" effects come into play as Wouter noted. This is usefully viewed by treating subsets as 0-1 vectors, and looking for sums and differences with net result a 0-1 vector. If these enumerations and earlier natural conjectures are true, we now know #(N) for N < 42. The techniques I used (and haven't elaborated) appear to generalize to any product of 3 primes, but hand calculation quickly becomes unfeasible. Handling a product of 4 primes looks MUCH harder.
From: Phil Carmody <thefatphil@yahoo.co.uk> (for N = 30) ... {0,1,7,13,19,20} is the symmetry-less set with minimal largest member that cancels.
{0,1,7,13,19,20} lifts to multiple of 30, but it is symmetric about line {9,24}. Adding the right balanced doubleton might give the reverse-lexicographically smallest set satisfying your property. Best, - Scott