The OEIS sequence A051772 defines the "distended" numbers as those positive integers $n$ for which each divisor of $n$ is greater than the sum of all smaller divisors.
Now, here's a question about such numbers:
Question: Are the "distended" numbers precisely the positive integers $n$ for which the "sum" function from subsets of the divisors of $n$ to the nonnegative integers is injective?
It is easy to show that any "distended" number $n$ satisfies the injectivity condition. Indeed, just write the divisors of $n$ in order from greatest to least as columns and below those columns, write all the binary numbers whose number of digits is at most the number of divisors of $n$, with leading zeros for binary numbers with fewer digits. Then, write a "sum" column that lists the sum of the divisors corresponding to the positions of the digit $1$ for each binary number. Finally, observe that each sum is less than the one below it because $n$ is a "distended" number. In particular, the "sum" column cannot contain any duplicates, so injectivity is satisfied.
What I could not prove is the converse. Perhaps, there might be a deficient counterexample for the converse. Also, any abundant counterexample for the converse would have to be a weird number, for injectivity would imply that the number could not be pseudoperfect. Of course, no abundant number could be a "distended" number.
There are lots of counterexamples.
Let $p$ be a prime. Then each prime $q$ between $p^2 - p-1$ and $p^2$, with at most $2016$ exceptions, produces a counterexample $n = p^2 q$.
Proof:
The divisors of $n = p^2 q$ are $1, p, q, p^2, pq, p^2 q$, in that order if $q$ is in the given interval. Since $p^2 < 1 + p + q$, $n$ is not distended.
Now there are $64 \cdot 63/2 = 2016$ unordered pairs of distinct sets of divisors. The sum of the first set minus the sum of the second corresponds to a polynomial of degree $\le 1$ in $q$, which must be $0$ if these two sets have the same sum. The polynomial is never identically $0$, so there is at most one $q$ value where the two sets have the same sum. Thus, for each $p$, there are at most $2016$ exceptional $q$.
Hmm, too bad this didn't come up $4$ years ago.