Consider the sum of $k^{th}$-power of divisors of $n$, denoted $$\sigma_k(n)=\sum_{d\vert n}d^k.$$ Let $\nu_p(x)$ stand for the $p$-adic valuation of the integer $x$.
The following appears to be true but is it?
Question: Fix $k, \ell\in\mathbb{N}$. If $k$ and $\ell$ have the same parity then $$\nu_2(\sigma_k(n))=\nu_2(\sigma_{\ell}(n)).$$
As noted in the comment by user44191, one needs only check this for prime powers. Note that this is trivial for $q=2$ and so we may assume that we have an odd prime $q$. Then the claim is that when $k \equiv \ell$ one has that $$v_2(\sigma_k(q^m)) = v_2\sigma_\ell (q^m).$$
This is the same as asserting that $$v_2(1+q^{k} + q^{2k} \cdots q^{mk}) = v_2(1+q^{\ell} + q^{2\ell} \cdots q^{m\ell}) $$
or equivalently that $$v_2\left( \frac{(q^k)^m-1}{q^k-1}\right) = v_2\left( \frac{(q^\ell)^m-1}{q^\ell-1}\right). $$
And that's the same as $$v_2\left( \frac{(q^m)^k-1}{q^k-1}\right) = v_2\left( \frac{(q^m)^\ell-1}{q^\ell-1}\right). $$
But this is true not just for prime $q$ but in fact for all odd $q$ because $\phi(2^n)$ is always a power of 2, and if $x \equiv 1$ (mod $2^n$) then $x^2 \equiv 1$ (mod $2^{n+1}$).
