Denote $A_k := \{\sigma : |I(\sigma)| = k\}$. Prove $\sum_{\sigma \in A_k} \sum_ {i \in I(\sigma)} (i+ \sigma(i)) = (n+1)k|A_k|$. Then the required sum will be $\sum_{k=1}^n (n+1)|A_k| = (n+1)!$. Or check the RMC 2014.

Ask Aiscrim.

Think of a $ \sigma '\in S_n $ such that $ \sigma '(n+1-\sigma (i))=n+1-i $. That kills it

Draw $n\times n$ grid in the plane with the lower left in $(1,1)$ and the upper right in $(n+1,n+1)$.Now start marking points $(i,\sigma(i))$,they're clearly all inside the gird.Note that $I(\sigma)$ is exactly those points on the left side when girded is divided by the main diagonal. Adjoin to permutation $\sigma$ a new one $\sigma'$ by reflecting the grid over the other diagonal.Note that :$$\sum_{i\in I(\sigma)} (i+\sigma(i))+\sum_{i\in I(\sigma')} (i+\sigma'(i))=2I(\sigma)(n+1)$$as $I(\sigma)=I(\sigma')$.Now multiplying the above sum by two and pairing off $(\sigma,\sigma')$ one obtains that the sum is actually $\sum_{\sigma \in S_n} n+1=(n+1)!$.$\blacksquare$ Just a remark,if one looks at the incidence matrix of the permutation the $(i+\sigma(i))$ becomes the sum of sides of a rectangle made out of $(i,\sigma(i))$ and its projections on the axes which seemed reasonably nice ,but i couldn't force a solution out of it.