If $S(n)$ is an odd integer, what is the sum of all possible $\frac1n?$ If $n$  is a positive integer, let $S(n)$ be the sum of all the positive divisors of $n$.
If $S(n)$ is an odd integer, what is the sum of all possible $\frac1n?$
The function $S$ is multiplicative and so, if we have the prime factorisation $n = p_1^{a(1)}p_2^{a(2)} \cdots p_m^{a(m)}$, where $p_1,p_2,...,p_m$ are distinct primes, then how will I continue to solve this?

$S(n)$ is multiplicative, so $S(n) = \prod_{i=1}^m S(p_i^{a_i})$. If any of the $S(p_i^{a_i})$ are even, the whole product will be even. So what is the condition on $S(p^a)$ being even for some prime $p$ and natural number $a$? What does that tell you about the set of numbers $n$ for which $S(n)$ is odd?
Once you've figured this out, you're going to need to sum the series. You'll want to factor out a geometric series involving powers of $2$, and you'll need to use the fact that $\sum_{k=0}^\infty k^{-2}  = \pi^2/6$.
