Odd permutations $\tau\in S_n$ with $\sum_{k=1}^nk\tau(k)$ a square

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 a square?

Let $ a(n)$ denote the number of odd permutations $ \tau\in S_n$ with $ \sum_{k=1}^nk\tau(k)$ a square. Via a computer I find that $ $ (a(1),\ldots,a(11))=(0,1,0,2,8,22,82,718,4583,53385,513897).$ $ 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!