This is an elaboration of a now deleted comment, written at the request of OP. I've had a couple of conjectures published on the OEIS which I really think justify some obvious numerical observations one can make about perfect numbers, harmonic divisor (Ore) numbers, and multiply-perfect numbers, whose definitions follow.
Let $\sigma(n)$ denote the sum of divisors of a positive integer $n$, and let $\sigma_0(n)$ denote the number of divisors of $n$.
$n$ is a harmonic divisor number iff $\sigma(n) \mid n\sigma_0(n)$
$n$ is multiply-perfect iff $n \mid \sigma(n)$.
$n$ is perfect iff $\sigma(n)=2n$. Every perfect number is obviously multiply-perfect, and Ore proved that every perfect number is a harmonic divisor number, which is equivalent to showing that a perfect number is never a square.
The only known odd multiply-perfect number is $1$, and the same goes for the harmonic divisor numbers. It's been conjectured that there are no further odd terms in either sequence, and earlier came the conjecture that there are no odd perfect numbers. A further sequence, the practical numbers, is required for our conjectures.
Let $p_1^{a_1}p_2^{a_2}...p_{\omega(n)}^{a_{\omega(n)}}$ be the canonical prime factorization of $n$, where $\omega(n)$ is the number of distinct divisors of $n$. Also let $p_0=a_0=1$.
$n$ is practical iff $p_i^{a_i+1}-1 \leq \sigma(p_0^{a_0}p_1^{a_1}...p_i^{a_i})$ for every $i \in [0,\omega(n)]$. This is Stewart's structure theorem, and for $i=1$, dividing both sides by $\sigma(p_1^{a_1})$ shows that $p_1\leq 2$ i.e. $n$ is even for $\omega(n)>0$, which is every $n>1$ by the fundamental theorem of arithmetic. Alternatively, $n$ is practical iff every positive integer less than $n$ has a representation as a sum of distinct divisors of $n$. It's easy to see that every practical number greater than $1$ is even, since $2$ has no representation as a sum of distinct positive integers not including itself.
The conjectures then, which are based on numerical evidence alone, are that every harmonic divisor number and every multiply-perfect number is a practical number. Part of the appeal of which, to me, is that now we don't have make the exception for $n=1$ in both of the weak conjectures with no real justification for why this would be the only odd term. These conjectures are refinements of the previous conjectures, but not of the odd perfect number conjecture specifically - since the form of every even perfect number is already known and they are easily shown to be practical, and as we've already shown in two different ways, every practical number greater than $1$ is even - though each implies the conjecture.
Of course, this is no more solid of evidence than that for the old conjectures, but seeing as how those conjectures have been perpetuated based mostly on numerical evidence, we should try looking more closely at the congruences of these numbers than just their parity. These conjectures, by the criterion of Srinivasan - every practical number greater than $2$ is divisible by $4$ or $6$ - say you will not find any terms of the form $12m\pm 2$ (since we can verify $2$ belongs to neither sequence). More generally, it is known that $n=2^a$, where $a$ is a positive integer, is neither a multiply-perfect number nor a harmonic divisor number. It is also known that every practical number not of this form and not $1$, is of the form $n=2^apm$, where $p$ is a prime such that $2^a<p<2^{a+1}$ and $a,m$ are positive integers. Thus, for every $n>1$, under our conjectures, a necessary condition for $n$ to be a multiply-perfect or harmonic divisor number is to have a divisor $d$ of the form $d=2^ap$ for some odd prime $p$ and integer $a$ such that $2^a<p<2^{a+1}$. Every integer with the form of $d$ is a practical number, and this includes the even perfect numbers.
Examining the prime factorizations of the harmonic divisor numbers or multiply-perfect numbers you would notice a tendency towards having a relatively large portion of small prime factors. This is a general characteristic of practical numbers, a sort of dual to the definition of a prime. You would also find that this imbalance seems exaggerated in further special cases of the equality $\sigma(n)=kn/2^r$, which is to say it's likely true that all of these numbers are practical. They can equivalently be defined as $n$ such that the odd part of $n$ divides $\sigma(n)$. I know the harmonic divisor numbers, multiply-perfect numbers, and this last generalization of multiply-perfect numbers are all practical for hundreds of terms at least, but I've confirmed the multiply-perfect numbers furthest - for $5261$ terms with $\sigma(n)>2n$ - from the factorizations provided in Achim Flammenkamp's database. It's very quick and straightforward from the canonical prime factorization of $n$ and Stewart's inequality to determine if $n$ is practical.
Not many mathematicians have written about practical numbers. A large portion of the papers in which they appear are linked here.
To summarize, if every every positive integer $n$ such that the odd part of $n$ divides $\sigma(n)$ is a practical number, then every multiply-perfect number is a practical number. If every multiply-perfect number or every harmonic divisor number is a practical number, then every perfect number is even.
Edit: I've found quite an incredible generalization of the previous conjectures. Let $n$ be a positive integer such that $\dfrac{\sigma(n)}n=\dfrac{k}m$, where $\gcd(k,m)==1$ and $m$ is a practical number. We conjecture that $n$ must also be practical.