For any positive integer $n$, as usual we let $S_n$ be the symmetric group of all the permutations of $\{1,\ldots,n\}$.

**QUESTION**: Is it true that for each integer $n>3$ there is an odd permutation $\tau\in S_n$ such that $\sum_{k=1}^n k\tau(k)$ is an odd square?

Let $a(n)$ denote the number of odd permutations $\tau\in S_n$ with $\sum_{k=1}^nk\tau(k)$ an odd square. Via a computer I find that $$(a(1),\ldots,a(11))=(0,0,0,2,4,10,42,436,2055,26454,263040).$$ For example, $(2,4,1,3)$ and $(3,1,4,2)$ are the only two odd permutations in $S_4$ meeting our requirement; in fact, $$1\cdot2+2\cdot4+3\cdot1+4\cdot3=5^2\ \ \text{and}\ \ 1\cdot3+2\cdot1+3\cdot4+4\cdot2=5^2.$$ For $n=2,3,4,5$, there is no even permutation $\tau\in S_n$ such that $\sum_{k=1}^nk\tau(k)$ is a square.

I conjecture that the question has a positive answer. Any comments are welcome!