If such an arrangement exists, $\textstyle\text{Total}=n\sum\limits_{k=1}^nk\equiv\sum\limits_{k=1}^nk\pmod n$, then $n\mid\tfrac{(n-1)n(n+1)}2$, so ${}n$ is odd. If ${}n$ is odd, consider the following construction, where $n=2k+1$. \[ \begin{array}{cccc|cccc} 2&3&\cdots&k+2&k+1&k+2&\cdots&2k\\ 3&4&\cdots&k+3&k&k+1&\cdots&2k-1\\ \vdots&\vdots&\ddots&\vdots&\vdots&\vdots&\ddots&\vdots\\ 2k&2k+1&\cdots&k-1&k+2&k+3&\cdots&2\\ 2k+1&1&\cdots&k&k+3&k+4&\cdots&1\\\hline 2k&2k-1&\cdots&k&k-1&k-2&\cdots&2k+1 \end{array} \]