AoPS - Problem

It seems to be a very weak bound? Since $m$ itself is one of the divisors, the equal sum must be at least $m$. In particular, the total sum of divisors must be at least $n \cdot m$. On the other hand, since all divisors are of the form $\frac{m}{k}$ for some $1 \le k \le m$, the total sum of divisors is at most \[m+\frac{m}{2}+\frac{m}{3}+\dots+\frac{m}{m}=mH_m.\]Hence we must have $n \le H_m \sim \log m$ and hence $m \gg e^n$. In fact, one can show that the sum of divisors of $m$ is always bounded by $Cm\log\log m$ and hence $n \ll \log\log m$ so that in fact $m \gg e^{e^{Cn}}$.