I have reached the conclusion that the non-existence of odd perfect numbers is related to the following
Conjecture Let it be $R=\{d_1,d_2,...,d_n\}$ the set of distinct proper divisors less than $\sqrt{O}$ of some odd positive integer $O$, such that $d_1<d_2<...<d_n$, and such that for some divisor $d_k\in R$ holds that $d_k\leq d_n<3d_k$. Then, it exist no solution to the egyptian fraction $$\frac{2}{d_k}+\frac{1}{d_1}+\frac{1}{d_2}+...=1$$
The process to reach the conjecture is the following (for you to check it is correct):
Let $P$ be some odd perfect number.
Let $R=\left\{ d_{1},d_{2},...,d_{n}\right\}$ be the set of proper divisors of $P$ less than $\sqrt{P}$ excluding $1$, and $T=\left\{ \frac{P}{d_{1}},\frac{P}{d_{2}},...\frac{P}{d_{n}}\right\}$ be the set of proper divisors of $P$ greater than $\sqrt{P}$ excluding $P$.
As $P$ is a perfect number,
$$1+d_{1}+d_{2}+...+d_{n}+\frac{P}{d_{n}}+...+\frac{P}{d_{2}}+\frac{P}{d_{1}}=P (1)$$
Operating,
$$1+d_{1}+d_{2}+...+d_{n}=P-\frac{P}{d_{1}}-\frac{P}{d_{2}}-...-\frac{P}{d_{n}}$$
$$1+d_{1}+d_{2}+...+d_{n}=P\left(1-\frac{1}{d_{1}}-\frac{1}{d_{2}}-...-\frac{1}{d_{n}}\right)$$
$$\frac{1+d_{1}+d_{2}+...+d_{n}}{1-\frac{1}{d_{1}}-\frac{1}{d_{2}}-...-\frac{1}{d_{n}}}=P$$
As $1+d_{1}+d_{2}+...+d_{n}$ is an integer, it follows that $\frac{1}{1-\frac{1}{d_{1}}-\frac{1}{d_{2}}-...-\frac{1}{d_{n}}}$ must be integer. Therefore, $P$ is a perfect number only if both $1+d_{1}+d_{2}+...+d_{n}$ and $\frac{1}{1-\frac{1}{d_{1}}-\frac{1}{d_{2}}-...-\frac{1}{d_{n}}}$ are proper divisors of $P$.
It is trivial to show that one of the two expression is less than $\sqrt{P}$, and therefore belongs to $R$, and the other is greater than $\sqrt{P}$ and belongs to $T$.
It is clear that $1+d_{1}+d_{2}+...+d_{n}$ is greater than the greatest element of $R$, so subsequently we can state that
$$\left(\frac{1}{1-\frac{1}{d_{1}}-\frac{1}{d_{2}}-...-\frac{1}{d_{n}}}\right)=d_{k}\in R (2)$$
$$1+d_{1}+d_{2}+...+d_{n}=\frac{P}{d_{k}}\in T (3)$$
The proof showing that $d_k\leq d_n<3d_k$ is a bit long, so I skip it for the shake of briefness. And operating with $(2)$, the conjecture follows. I have found (thanks to some help provided here in MathStackExchange) examples of odd positive integers for which $(3)$ alone holds, but not any example of an odd positive integer for which $(2)$ holds with the restrictions exposed.
Any help on how to attack this conjecture would be welcomed. Otherwise, if someone is able to show any counterexample to the conjecture (that is, provide some odd positive integer for which $(1)$ holds), it would be more than welcomed to save efforts proving it. I have not seen this conjecture elsewhere, so it is possible that it is mistaken.
Thanks in advance!
Edit
I post the proof showing that $d_k\leq d_n<3d_k$, as it is fundamental for the conjecture and for the solution I have provided for verification.
From (3), and operating,
$$1+\sum_{j=1}^{n}d_{j}=\frac{P}{d_{k}}$$
$$P=d_{k}\left(1+\sum_{j=1}^{n}d_{j}\right)$$
As by definition we have that $d_{n}<\sqrt{P}$, just substituting we can set that
$$d_{n}<\sqrt{d_{k}\left(1+\sum_{j=1}^{n}d_{j}\right)}$$
$$d_{n}^{2}<d_{k}\left(1+\sum_{j=1}^{n}d_{j}\right)$$
$$\frac{d_{n}}{d_{k}}<\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}} (4)$$
Other hand, taking (1) and dividing by $\sqrt{P}$, we get that
$$\frac{1}{\sqrt{P}}+\frac{d_{1}}{\sqrt{P}}+\frac{d_{2}}{\sqrt{P}}+...+\frac{d_{n}}{\sqrt{P}}+\frac{P}{d_{n}\sqrt{P}}+...+\frac{P}{d_{2}\sqrt{P}}+\frac{P}{d_{1}\sqrt{P}}=\sqrt{P}$$
$$\frac{1}{\sqrt{P}}+\frac{d_{1}}{\sqrt{P}}+\frac{d_{2}}{\sqrt{P}}+...+\frac{d_{n}}{\sqrt{P}}+\frac{\sqrt{P}}{d_{n}}+...+\frac{\sqrt{P}}{d_{2}}+\frac{\sqrt{P}}{d_{1}}=\sqrt{P}$$
As $\forall d_{j}\,\frac{\sqrt{P}}{d_{j}}>1$, and $1+\sum_{j=1}^{n}d_{j}>\sqrt{P}$, then we get inmediately that $n<\sqrt{P}-1$; otherwise, the sum of all the terms would be greater than $\sqrt{P}$ and $P$ would not be a perfect number. Thus, we can affirm that
$$n\leq\sqrt{P}-2 (5)$$
As the minimum gap between consecutive elements of $R$ is $2$, the maximum sum of elements of $R$ with the minimum gap between them is
$$d_{n}+\left(d_{n}-2\right)+\left(d_{n}-4\right)+...+\left(d_{n}-2\left(n-1\right)\right)$$
Therefore, we can establish that
$$1+\sum_{j=1}^{n}d_{j}\leq d_{n}+\left(d_{n}-2\right)+\left(d_{n}-4\right)+...+\left(d_{n}-2\left(n-1\right)\right)$$
$$1+\sum_{j=1}^{n}d_{j}\leq nd_{n}-\left(\left(n-1\right)^{2}+\left(n-1\right)\right)$$
Subsequently, we get that
$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}\leq\frac{nd_{n}-\left(\left(n-1\right)^{2}+\left(n-1\right)\right)}{d_{n}}$$
$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}\leq n-\frac{\left(\left(n-1\right)^{2}+\left(n-1\right)\right)}{d_{n}}$$
Substituting $n$ by the inequality obtained in (5), we get that
$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}\leq\sqrt{P}-2-\frac{\left(\left(\sqrt{P}-2-1\right)^{2}+\left(\sqrt{P}-2-1\right)\right)}{d_{n}}$$
$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}\leq\sqrt{P}-2-\frac{\left(\left(\sqrt{P}-3\right)^{2}+\left(\sqrt{P}-3\right)\right)}{d_{n}} (6)$$
Operating with the numerator of the third term of the right handside of (6), we get that
$$\left(\sqrt{P}-3\right)^{2}+\left(\sqrt{P}-3\right)=$$
$$P+9-6\sqrt{P}+\sqrt{P}-3=$$
$$P-5\sqrt{P}+6$$
As by definition $d_{n}<\sqrt{P}$, we can affirm that
$$\frac{P-5\sqrt{P}+6}{d_{n}}>\sqrt{P}-5+\frac{6}{\sqrt{P}}$$
Therefore, substituting at (6), we get that
$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}<\sqrt{P}-2-\left(\sqrt{P}-5+\frac{6}{\sqrt{P}}\right)$$
$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}<3-\frac{6}{\sqrt{P}}$$
Subsequently,
$$\frac{1+\sum_{j=1}^{n}d_{j}}{d_{n}}<3$$
$$1+\sum_{j=1}^{n}d_{j}<3d_{n}$$
Finally, substituting in (4), we get the desired result
$$\frac{d_{n}}{d_{k}}<3$$