Let $\sigma(n)$ be the sum of the divisors of $n$. Is it always true that if $n$ is odd, that $$n\mid\sum_{k=1}^{\frac{n-1}{2}}k^2\sigma(k)\sigma(n-k)?$$
I have checked this up to $n=100$, and I suspect there's some simple argument for why this should be true, probably using Eisenstein series identities but I'm not seeing it.
The motivation from this is from the old Mathoverflow question Van der Pol's identity for the sum of divisors and a quartic polynomial equation for odd perfect numbers, where this term appears as the constant in the polynomials generated which for an integer $n$, $n$ is an odd perfect number if and only if $n$ is a root of the $n$th polynomial there. It seems unlikely that the method used there will work as any sort of viable attack on that problem, since they have a very tough set of quartic polynomials they would need to prove are irreducible, and one has very little control over the behavior of the coefficients. But if the above relationship is true, then this at least can be replaced with a similar looking cubic polynomial which still seems unlikely to be helpful, but at least marginally more plausible as an angle of productive attack since then if there is an integer root, the other two roots have to be quadratics. And the coefficient of the $x^2$ will always be $-1$, so the sum of the roots is pretty restricted.
Best Answer
From the paper of Touchard that is linked in the question on we get the relation $$3nS_0(n)-\frac{n(n-1)\sigma(n)}{6}=\frac{10}{n}S_2(n) ....(1)$$
here $S_i(n)=\sum_{k=1}^{n-1}k^i\sigma(k)\sigma(n-k)$.
Now we can prove $n\nmid S_2(n)$ in general when $n$ is even. For $n$ odd, first take the case when $5\nmid n$.
We need to prove $6|n(n-1)\sigma(n)$. If $3|n$ we are done ($2$ divides $n-1$). Otherwise: we have $n=\prod p_i^{a_i}, 3\nmid p$. Now $n \not\equiv 1 (\mod 3)$ when odd number primes like $p_k \equiv -1 (\mod 3)$ have powers $a_k$ that are odd. But $\sigma(n)=\prod_{i} \frac{p_i^{a_i+1}-1}{p_i-1}$. So, for those primes $p_k \equiv -1 (\mod 3)$, $p_k^{a_k+1}\equiv 1 (\mod 3)$ as $a_k$ is odd and $3 \nmid p_k-1$. So, we get that $6|n(n-1)\sigma(n)$.
Now, for the case $5\mid n$,we divide both sides by $5$. We are left to prove that $30\mid n(n-1)\sigma_(n)$. As $5\mid n$, we need to prove $6|m(5^am-1)\sigma(5^am)$. Here, $n=5^am, 5\nmid m$. Again, as $2|(n-1)$ we need to prove $3|m(5^am-1)\sigma(5^am-1)$. We focus on the case when $3 \nmid m$, because the other case is trivial. Again $5^am \not\equiv 1 (\mod 3) $ when odd number of primes $p_k$ s.t $p_k \equiv -1 (\mod 3)$ (including $5$) of $m$ have odd powers $a_k$, so there's at least such prime. We do the same thing as previous case and find that $3|\sigma(5^am)$ which proves that $n|S_2(n)$.