Let elements in $S$ are $a_1<a_2<...<a_k$ and $k \geq 3$ We have $\frac{a_1^2}{a_k-a_1} \geq a_1 \to a_k \leq 2a_1$ $\frac{a_{k-1}^2}{a_k-a_{k-1}} \in S $ and $\frac{a_{k-1}^2}{a_k-a_{k-1}}>\frac{a_{k-1}^2}{2a_{k-1}-a_{k-1}}=a_{k-1}$ So $\frac{a_{k-1}^2}{a_k-a_{k-1}}=a_k \to a_k^2-a_ka_{k-1}-a_{k-1}^2=0 \to \frac{a_k}{a_{k-1}}=\frac{1+\sqrt{5}}{2}$ - irrational So only two elements is possible And we can prove that $\frac{a_1^2}{a_2-a_1}=a_2$ has not integer solutions, so $\frac{a_1^2}{a_2-a_1}=a_1 \to a_2=2a_1$ So all sets are $(n,2n)$

Baltic Way 2008 P6

Thanks for the solution! I just read about the Olympic tasks. My brother thinks they are too easy to compete. And that everyone who unleashes for a while can easily be mistaken from a finite set in alignment.

YOU NEED TO GET THE LITTLE TWO. $a,2a$