APMO

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 $\ell >2⟹\frac{\ell }{2}+1>2$, $\frac{\ell }{2}+1\ge \ell ⟺\ell \le 2$, a contradiction again.

Therefore $S$ can only contain $2$, and $S=\{2\}$ is the only solution.