Five With Zero Sum

Worked solution

The answer is 9. Lower bound: the eight integers 0, 0, 0, 0, 1, 1, 1, 1 contain no good five. Any five of them use $j$ ones with $1 \le j \le 4$, giving sum $j$, never divisible by 5. So $N \ge 9$. Upper bound: this is the Erdős-Ginzburg-Ziv theorem, which says any $2n - 1$ integers contain $n$ whose sum is divisible by $n$. For prime $n = p$ a slick proof uses the Chevalley-Warning theorem, or Cauchy-Davenport. For $n = 5$ only residues matter, and there are just $\binom{13}{4} = 715$ multisets of 9 residues mod 5, so a direct check that every one contains a zero-sum 5-subset also settles it. Hence $N = 9$.