Let $\sigma(x)$ be the sum of the divisors of $x$. A number $X$ is called perfect if $\sigma(X) = 2X$. Denote the abundancy index $\sigma(X)/X$ by $I(X)$.
If $N$ is odd and perfect, then $N$ can be written in the Eulerian form $N = {q^k}{n^2}$, where $q$ is prime with $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q, n) = 1$.
Here is my question:
If $N = {q^k}{n^2}$ is an odd perfect number given in Eulerian form, does $\sigma(n^2)/q$ divide $2n^2 - \sigma(n^2)$?
Context (as requested by Hagen von Eitzen)
Since $N = {q^k}{n^2}$ is perfect, we have $$2{q^k}{n^2} = 2N = \sigma(N) = \sigma(q^k)\sigma(n^2).$$
Now write $\sigma(q^k)$ as $$\sigma(q^k) = q^k + \sigma(q^{k-1})$$ so that $${q^k}\left(2n^2 - \sigma(n^2)\right) = \sigma(q^{k-1})\sigma(n^2).$$ Since $\gcd\left(q,\sigma(q^{k-1})\right) = 1$, then $q \mid \sigma(n^2)$.
Consequently, we obtain $$I(q^{k-1}) = \frac{2n^2 - \sigma(n^2)}{\sigma(n^2)/q}.$$
Thus, if $k \geq 5$, $$1 < \frac{2n^2 - \sigma(n^2)}{\sigma(n^2)/q} = I(q^{k-1}) < \frac{5}{4}$$ so that $\sigma(n^2)/q \nmid \left(2n^2 - \sigma(n^2)\right)$.
By the contrapositive, $\sigma(n^2)/q \mid \left(2n^2 - \sigma(n^2)\right)$ implies the Descartes-Frenicle-Sorli conjecture that $k = 1$.
Observe that $k = 1$ implies that $$1 = I(1) = I(q^{k-1}) = \frac{2n^2 - \sigma(n^2)}{\sigma(n^2)/q}$$ so that $\sigma(n^2)/q \mid \left(2n^2 - \sigma(n^2)\right)$.
Thus, the biconditional $$k = 1 \Longleftrightarrow \sigma(n^2)/q \mid \left(2n^2 - \sigma(n^2)\right)$$ holds, as already pointed out by Hagen von Eitzen in his answer.
As $\sigma(n^2)$ is a multiple of $ \frac{\sigma(n^2)}{q}$, the desired claim is equivalent to $\frac{\sigma(n^2)}{q}\mid 2n^2$. We have $2q^kn^2=\sigma(N)=(1+q+q^2+\ldots+q^k)\sigma(n^2)$, hence $$ \frac{\sigma(n^2)}{q}=\frac{2n^2q^{k-1}}{1+q+q^2+\ldots+q^k}$$ and we ask whether $$\frac{1+q+q^2+\ldots+q^k}{q^{k-1}}=q+1+q^{-1}+\ldots+q^{1-k}$$ is an integer. Quite apparently, this is the case if and only $k=1$.