Determine all finite nonempty sets $S$ of positive integers satisfying

$$
\frac{i+j}{(i, j)} \text { is an element of } S \text { for all } i, j \text { in } S \text {, }
$$

where $(i, j)$ is the greatest common divisor of $i$ and $j$.
Answer: $S=\{2\}$.
Let $k \in S$. Then $\frac{k+k}{(k, k)}=2$ is in $S$ as well.
Suppose for the sake of contradiction that there is an odd number in $S$, and let $k$ be the largest such odd number. Since $(k, 2)=1, \frac{k+2}{(k, 2)}=k+2>k$ is in $S$ as well, a contradiction. Hence $S$ has no odd numbers.
Now suppose that $\ell>2$ is the second smallest number in $S$. Then $\ell$ is even and $\frac{\ell+2}{(\ell, 2)}=\frac{\ell}{2}+1$ is in $S$. Since $\ell>2 \Longrightarrow \frac{\ell}{2}+1>2, \frac{\ell}{2}+1 \geq \ell \Longleftrightarrow \ell \leq 2$, a contradiction again.
Therefore $S$ can only contain 2, and $S=\{2\}$ is the only solution.