Solution:

Let the subsets be $S_{1}, S_{2}, \ldots, S_{k}$. Suppose the statement is false and seek a contradiction. Then for each $S_{i}$ there exists some $n_{i}$ such that $S_{i}$ contains only finitely many multiples of $n_{i}$. Let $n = n_{1} n_{2} \cdots n_{k}$; then every multiple of $n$ is a multiple of each $n_{i}$ and so each $S_{i}$ can contain only finitely many multiples of $n$. But this means that the $k$ sets together contain only finitely many multiples of $n$, and since they partition the positive integers (which contain infinitely many multiples of $n$), we have our contradiction.