Switzerland — Final Round

Solution:
Trying to apply number theoretical methods to deduce something from the fact that $m-n$ divides $n^{2}$ does not seem to lead anywhere. Instead, we will try to find the extreme values that the quotient $n^{2} /(m-n)$ can achieve. This will give us some interesting bounds on the elements of $S$.

a. First, since $S$ contains at least two elements, we take $l > s$ to be the largest and smallest elements of $S$. Since $s^{2} /(l-s)$ also belongs to $S$ we must have
$$\frac{s^{2}}{l-s} \geq s \quad \Longrightarrow \quad 2s \geq l$$

b. Now, consider $k < l$ the second largest element of $S$. The number $k^{2} /(l-k)$ belongs to $S$. We claim that $k^{2} /(l-k) \neq l$, which implies
$$\frac{k^{2}}{l-k} \leq k \quad \Longrightarrow \quad 2k \leq l$$
We prove the previous claim in the following lemma.

Lemma 1
The equation $k^{2} /(l-k) = l$ has no positive integer solutions $l > k$.

Proof. Assume such a solution exists, then it also solves
$$l^{2} = k^{2} + k l$$
Since the equation is homogeneous (of degree 2), we can assume that $\operatorname{gcd}(k, l) = 1$. Since we must have $l \mid k^{2}$, we deduce that $l = 1$. This is a contradiction since $k$ is a positive integer with $k < l$.

Combining the two relations above, and using that $s \leq k$, we get
$$l \leq 2s \leq 2k \leq l$$
Therefore $k = s$ and $S$ contains exactly two elements. Moreover, the inequalities above also imply $l = 2s$. So $S = \{s, 2s\}$ as desired.

Any such set satisfies the desired property because
$$\frac{s^{2}}{2s - s} = s \in S$$