I was reading a paper, and it claimed that only the divisors of an odd abundant number would satisfy this. Keep in mind that all divisors need not be used. There doesn't seem to be a proof given, so is this true? And how would one prove it?
Paper's link: https://www.sciencedirect.com/science/article/pii/0012365X73901362
On
When you say that all of the divisors don't have be used, I believe what you are making reference to are pseudoperfect numbers; i.e., positive integers that are equal to the sum of some of their divisors.
For example, consider
$$ 945 = 315 + 189 + 135 + 105 + 63 + 45 + 35 + 27 + 15 + 9 + 7 .$$
Now, if we divide both sides by 945, we get
$$ 1 = \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \frac{1}{9} + \frac{1}{15} + \frac{1}{21} + \frac{1}{27} + \frac{1}{35} + \frac{1}{63} \frac{1}{105} + \frac{1}{135}. $$
It's not hard to see that this will happen with every pseudoperfect number.
For the base case of $x_1 = 1$ and $k = 1$, note $1$ is a divisor of all positive integers, including each odd Abundant number. With $k \ge 2$, it's fairly easy to show that all $x_i$ must also be divisors of an odd abundant number. We're given
$$\sum_{i=1}^k \frac{1}{x_i} = 1 \tag{1}\label{eq1}$$
where $x_i \lt x_j$ for all $i \lt j$ and all $x_i$ are odd integers. If the lcm of the $x_i$ is not an odd perfect number or they don't include all factors of the lcm apart from $1$, then I can multiply both sides by the lcm value to prove the result. However, although there are no known odd perfect numbers (Odd perfect numbers says any such number must be $\gt 10^{1500}$, have at least $101$ prime factors and at least $10$ distinct prime factors, etc.), just in case, plus to keep things simpler, multiply both sides by $\prod_{j=1}^k x_j$ instead to get
$$\sum_{i=1}^k \left(\prod_{j=1,j \neq i}^k x_j\right) = \prod_{j=1}^k x_j = a \tag{2}\label{eq2}$$
Since $x_i$ times each $i$'th left side term gives $a$, and $x_i \neq x_j$ for $i \neq j$, each left side term is a unique factor of $a$. As the terms sum to $a$, but don't include all of the factors (e.g., $1$ and $a$ itself are not included), this means the sum of all of the factors of $a$ must be larger than itself, i.e., $a$ is an odd abundant number, with the odd part coming from all $x_i$ being odd.
Update: As Gerry Myerson stated in the question comments, and as I explained further, all finite sets of odd integers are a set of divisors of an odd abundant number, with this being due to there being at least one odd abundant number (e.g., $945$) and every multiple of an abundant number being abundant. Thus, perhaps the author was referring to a specific odd abundant number, with this most likely being either the lcm of the $x_i$ or the product of them, i.e., $a$ in \eqref{eq2}.