Subset sums

Simpler version. Among any $n+1$ integers, two have difference divisible by $n$ (as shown previously). Let these be $a,b$ with $b-a=kn$. Then the subset $\{a,b\}$... but that's a difference, not a sum.

Corrected version. Among any $n+1$ integers, there exist some whose sum is divisible by $n+1$: consider partial sums $S_k=a_1+\cdots+a_k$ for $k=1,\ldots,n+1$, mod $n+1$. If some $S_k\equiv 0$, done. Otherwise two $S_i,S_j$ have the same residue, and $S_j-S_i$ is a subset sum divisible by $n+1$.

Conclusion. A nonempty subset has sum divisible by $n+1$.