Alright, sorry for the delay; here's my solution.

Suppose \(\displaystyle \sigma\) is such that the sum on the right is minimal. Then we can assume \(\displaystyle \sigma\) to be a permutation of \(\displaystyle N=\{1, \dots, n\}\); indeed, order the values \(\displaystyle \sigma(1), \dots, \sigma(n)\) in increasing order and replace them respectively by \(\displaystyle 1, \dots, n\); we minimize the sum in such a way. Now suppose there exist \(\displaystyle j,k \in N\) with \(\displaystyle j < k\) and \(\displaystyle \sigma(j) > \sigma(k)\); suppose we switch the values of \(\displaystyle \sigma(j), \sigma(k)\) in the sum. The difference between the two sums is \(\displaystyle \left(\frac{\sigma(j)}{j^2}+\frac{\sigma(k)}{k^2}\right) - \left(\frac{\sigma(k)}{j^2}+\frac{\sigma(j)}{k^2}\right)>0\), meaning we have minimized the sum some more, contradicting the hypothesis. Therefore \(\displaystyle j<k \Rightarrow \sigma(j)<\sigma(k) \Rightarrow \sigma=\mbox{i.d.}_N\).