I believe statement should be: Determine all positive integers $n$ such that numbers from $1$ to $n$ can be sorted in some order $x_1,x_2,...,x_n$ with the property that the number $x_1+x_2+...+x_k$ is divisible by $k$, for all $1\le k\le n$. If so then: $n=1$ is obvious answer Let $n >1$ $n|(1+2+...+n)=\frac{n(n+1)}{2}$ so $n$ is odd and $n=2k-1, k>1$ $2k-2=n-1|x_1+...+x_{n-1}=(1+2+...+n)-x_n=(2k-1)k-x_n=(2k-2)k+k-x_n$ As $2k-2 >k-1 \geq k-x_n$ then $x_n=k$ $2k-3=n-2|x_1+...+x_{n-2}=(1+2+...+(2k-1))-x_{n-1}-x_n=k(2k-1)-k-x_{n-1}=(2k-3)k+k-x_{n-1}$ $k \neq x_{n-1}$ so $k-1 \geq k-x_{n-1} \geq 2k-3 \to k \leq 2$ So $k=2$ and $n=3$ For $n=3$ we can choose sequence $1,3,2$