I was solving this CSES question called Sum of Divisors and one of the solutions in the USACO Guide hints at this statement: The minimum value of $j > i$ such that $\lfloor n/j \rfloor$ < $\lfloor n/i \rfloor$ is $j = \lfloor n/q \rfloor + 1$, where $q = \lfloor n/i \rfloor$ for integers $j, i...