Consider $\mathcal U_m$ the set of the finite subsets of $\mathbb N$ which admit $m$ as a maximum. Obviously, every non empty finite subset of $\mathbb N$ belongs to one and only one $\mathcal U_m$. Moreover, the cardinal of $\mathcal U_m$ is $2^m$.
Consider $f:\mathbb N \to \mathbb N$ a non decreasing function.
It is clear that if $U$ is in $\mathcal U_m$, then $ f(m) \leq \sum_{a\in U} f(a) \leq (m+1)f(m) $ Thus $ \frac{2^m}{(m+1)f(m)} \leq \sum_{U\in \mathcal U_m} \frac{1}{\sum_{a\in U} f(a)} \leq \frac{2^m}{f(m)} $
Now, set $f(a) = a^n$ or $f(a) = a^a$ and you have your answer by summing over $m$.