nice Let $a_k=n$. Obviously $a_1+a_2+...+a_k$ should be different than $a_1+a_2+...+a_k+a_{k+1}$ but this isn't possible unless $k=1$. So we have $a_1=n$. Now look at the final summand: $a_1+a_2+...+a_n=1+2+3+...+n=\frac{n(n+1)}{2}$. So we must have $\frac{n(n+1)}{2} \ne a_1 = n =0 \ (mod \ n)$ so we must have $n$ even, but we can find a construction for $n$ even: $\{ n,1,n-1,2,n-2,3,...\}$