Let $\sigma(x)=\sigma_1(x)$ be the classical sum of divisors of the positive integer $x$. Denote the aliquot sum of $x$ by $s(x)=\sigma(x)-x$ and the deficiency of $x$ by $d(x)=2x-\sigma(x)$. Finally, denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$.
A number $P$ is said to be perfect if $\sigma(P)=2P$. On the other hand, if a number $A$ satisfies $\sigma(A)=2A-1$, then $A$ is said to be almost perfect. (The only known examples for $A$ are the powers of two, including $1$.) Lastly, a number $D$ is said to be deficient-perfect if the deficiency $d(D)$ of $D$ divides $D$.
Even Perfect Numbers
The three smallest examples for $P$ are $$6=2\cdot{3}={2^1}\cdot\left(2^2 - 1\right)$$ $$28=4\cdot{7}={2^2}\cdot\left(2^3 - 1\right)$$ $$496={16}\cdot{31}={2^4}\cdot\left(2^5 - 1\right).$$
The Euclid-Euler Theorem states that an even number $E$ is perfect if and only if $E$ has the form $$E={2^{p-1}}\cdot\left(2^p - 1\right) = \frac{M_p (M_p + 1)}{2},$$ where $M_p = 2^p - 1$ is a Mersenne prime. (Note that if $M_p$ is prime, then $p$ is also prime. The converse does not necessarily hold.)
Only $51$ Mersenne primes are known, corresponding to an equal number of (even) perfect numbers. (See this GIMPS page.) It is currently unknown whether there are finitely many such numbers $E$. It has been conjectured that there are infinitely many even perfect numbers.
Note that we have the equations $$1=\gcd\left(M_p,\frac{M_p + 1}{2}\right)=\sigma\left(\frac{M_p + 1}{2}\right)/M_p=\frac{M_p + 1}{\sigma(M_p)}=\frac{d\left(\frac{M_p + 1}{2}\right)}{s(M_p)}=\frac{2s\left(\frac{M_p + 1}{2}\right)}{d(M_p)}$$ so that $$d\left(\frac{M_p + 1}{2}\right)d(M_p)=2s\left(\frac{M_p + 1}{2}\right)s(M_p).$$
(Note that, trivially, we have $(M_p + 1)/2 < M_p$.)
Odd Perfect Numbers
Much less is known about odd perfect numbers $O$. We do not know an actual value for even just a single divisor of $O$, apart from the trivial factor $1$. This is because we do not have an example for $O$, unlike the case for $E$. (In a letter to Mersenne dated November $15$, $1638$, Descartes showed that $$\mathscr{D} = {{3}^2}\cdot{{7}^2}\cdot{{11}^2}\cdot{{13}^2}\cdot{22021} = 198585576189$$ would be an odd perfect number if $22021$ were prime.)
Nonetheless, Euler did prove that $O$ must necessarily have the so-called Eulerian form $$O = q^k n^2$$ where $q$ is the special prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Note that we have the equations $$1 < \gcd(n^2,\sigma(n^2))=\frac{\sigma(n^2)}{q^k}=\frac{2n^2}{\sigma(q^k)}=\frac{d(n^2)}{s(q^k)}=\frac{2s(n^2)}{d(q^k)} \tag{*}$$ so that $$d(q^k)d(n^2)=2s(q^k)s(n^2).$$
(It is known that $q^k < n^2$.)
Odd Perfect Numbers Must Have a Deficient-Perfect Divisor
Remark 1. We may define a deficient-perfect number $D$ to be a positive integer having a proper divisor $e$ satisfying $$\sigma(D) = 2D - e.$$ The divisor $e$ is said to be the deficient divisor of $D$.
Remark 2. It is also easy to show that the deficient divisor $e$ of a deficient-perfect number $D$ is the greatest common divisor of $D$ and $\sigma(D)$. That is, we have $$e=\gcd(D,\sigma(D))$$ if $σ(D) = 2D − e$ and $e | D$, $e \neq D$.
Holdener and Rachfal (2019) proved the following assertion:
THEOREM 1. Every odd perfect number has a deficient-perfect divisor. In particular, if $O = q^k n^2$ is an odd perfect number given in Eulerian form, then the divisor $D_O = q^j n^2$ is deficient-perfect, with deficient divisor $$e_O=\frac{2q^j n^2}{q^{j+1} + 1},$$ where $j=(k-1)/2$.
In particular, by Remark 1 and Remark 2, we obtain $$\gcd\left(D_O,\sigma(D_O)\right)=e_O=2D_O-\sigma(D_O)=d(D_O).$$
Following in the footsteps of this short ResearchGate article, titled "Gcd in Odd perfect number" by Devansh Singh (B.Tech, M.E., I.E.T. Lucknow), we proceed as follows:
From the Eulerian form $O=q^k n^2$, we get $$O={q^{(k+1)/2}}\cdot\left(q^{(k-1)/2} n^2\right)={q^{(k+1)/2}}\cdot{D_O}.$$ Since $\gcd\left(q^{(k+1)/2},q^{(k-1)/2} n^2\right)=1$ if and only if $k=1$, then we are unable to tie up, in full generality, the value of $$\sigma(O/q^k)/q^{j+1}=\sigma(n^2)/q^{(k+1)/2}$$ with that of $$\gcd(D_O,\sigma(D_O))=\gcd\left(q^{(k-1)/2} n^2,\sigma(q^{(k-1)/2} n^2)\right),$$ by using Singh's approach.
This is getting too long already, so let me ask my question (as is) in the title:
QUESTION. Does an odd perfect number have a divisor (other than $1$) which must necessarily be almost perfect?
MY ATTEMPT
Note from the definitions that, if a number is almost perfect, then it is automatically deficient-perfect.
Suppose to the contrary that the deficiency $d(D_O)$ of the deficient-perfect divisor $D_O=q^j n^2$ of an odd perfect number $O=q^k n^2$ satisfies $d(D_O)=1$ (where $j=(k-1)/2$). (Notice that $D_O$ is a square.)
In other words, assume that $D_O$ is almost perfect. Then we have, by a criterion in this paper, the following inequality $$\frac{2D_O}{D_O + 1}< I(D_O) < \frac{2D_O + 1}{D_O + 1}$$ since $D_O > 1$ and $D_O$ is almost perfect.
Using the fact that the abundancy index function $I$ is multiplicative, we obtain $$I(D_O)=I(q^j n^2)=I(q^j)I(n^2).$$
If $k=1$, then using the equations and inequality in $(*)$, we obtain $$1<\gcd(n^2,\sigma(n^2))=d(n^2),$$ which contradicts our earlier assumption that $d(D_O)=1$. Thus, it suffices to consider $k \neq 1$. Since $k \equiv 1 \pmod 4$ holds, then $k \geq 5$. It follows that $j=(k-1)/2 \geq 2$.
In particular, we have $$I\left((qn)^2\right) = I(q^2)I(n^2) \leq I(D_O) < \frac{2q^j n^2 + 1}{q^j n^2 + 1} \tag{1}$$ and $$\frac{2q^j n^2}{q^j n^2 + 1} < I(D_O) = I(q^j)I(n^2) = \left(\frac{q^{j+1} - 1}{q^j (q - 1)}\right)\cdot{I(n^2)} < \left(\frac{q}{q - 1}\right)\cdot\left(\frac{2q}{q + 1}\right) = \frac{2q^2}{q^2 - 1}. \tag{2}$$
Inequality $(2)$ results to $$0 < 2\left(q^j n^2 + q^2\right)$$ which is trivial.
On the other hand, we derive the following estimate from Inequality $(1)$: $$I(q^2)I(n^2)=\left(\frac{\sigma(q^2)}{q^2}\right)\cdot{I(n^2)}=\left(\frac{q^2 + q + 1}{q^2}\right)\cdot{I(n^2)}$$ $$>\left(\frac{q^2 + q + 1}{q^2}\right)\cdot\left(\frac{2(q-1)}{q} + \frac{1}{qn^2}\right).$$
We now test whether the inequality $$\left(\frac{q^2 + q + 1}{q^2}\right)\cdot\left(\frac{2(q-1)}{q} + \frac{1}{qn^2}\right) < \frac{2q^j n^2 + 1}{q^j n^2 + 1} \tag{3}$$ indeed holds. Using some help from WolframAlpha, Inequality $(3)$ holds when $q < n$.
To conclude, $D_O = q^j n^2$ might be an almost perfect divisor of the odd perfect number $O=q^k n^2$, where $j=(k-1)/2$, as we were not able to derive a contradiction from assuming that $D_O$ is indeed almost perfect, and then using our criterion for almost perfect numbers.
I will stop here for the time being, as I have got some more work to do.
Assume to the contrary that the deficient divisor $e_O$ satisfies $$1 = e_O = \frac{2q^j n^2}{q^{j+1} + 1}. \tag{1}$$ (Note that this condition holds if and only if $d(D_O)=2D_O-\sigma(D_O)=1$.)
At once, we see that $$(1) \iff q^{j+1} + 1 = 2q^j n^2$$ $$\iff q + \left(\frac{1}{q}\right)^j = 2n^2, \tag{2}$$ where one notices that the left-hand side of Equation $(2)$ is not an integer (since $j \geq 2$), which contradicts the fact that the right-hand side of Equation $(2)$ must be an integer.
Thus, $D_O$ is not almost perfect.
(I would like to qualify that I have not re-read Holdener and Rachfal's paper, so I am guessing their proof might be very similar to this one.)