Find all finite sets $S$ of positive integers with at least $2$ elements, such that if $m>n$ are two elements of $S$, then 
$$ \frac{n^2}{m-n} $$ 
is also an element of $S$. 
1. Let the elements of \( S \) be \( a_1 < a_2 < \ldots < a_k \) where \( k \geq 2 \). We need to find all such sets \( S \) such that for any \( m > n \) in \( S \), the value \( \frac{n^2}{m-n} \) is also an element of \( S \).

2. Consider the largest element \( a_k \) and the smallest element \( a_1 \) in \( S \). We have:
   \[
   \frac{a_1^2}{a_k - a_1} \in S
   \]
   Since \( \frac{a_1^2}{a_k - a_1} \) must be a positive integer and an element of \( S \), it must be at least \( a_1 \). Therefore:
   \[
   \frac{a_1^2}{a_k - a_1} \geq a_1
   \]
   Simplifying this inequality:
   \[
   \frac{a_1^2}{a_k - a_1} \geq a_1 \implies a_1^2 \geq a_1(a_k - a_1) \implies a_1^2 \geq a_1 a_k - a_1^2 \implies 2a_1^2 \geq a_1 a_k \implies a_k \leq 2a_1
   \]

3. Now consider the second largest element \( a_{k-1} \) and the largest element \( a_k \). We have:
   \[
   \frac{a_{k-1}^2}{a_k - a_{k-1}} \in S
   \]
   Since \( \frac{a_{k-1}^2}{a_k - a_{k-1}} \) must be greater than \( a_{k-1} \), we have:
   \[
   \frac{a_{k-1}^2}{a_k - a_{k-1}} > a_{k-1}
   \]
   Simplifying this inequality:
   \[
   \frac{a_{k-1}^2}{a_k - a_{k-1}} > a_{k-1} \implies a_{k-1}^2 > a_{k-1}(a_k - a_{k-1}) \implies a_{k-1}^2 > a_{k-1} a_k - a_{k-1}^2 \implies 2a_{k-1}^2 > a_{k-1} a_k \implies a_k < 2a_{k-1}
   \]

4. Suppose \( \frac{a_{k-1}^2}{a_k - a_{k-1}} = a_k \). Then:
   \[
   a_{k-1}^2 = a_k(a_k - a_{k-1}) \implies a_k^2 - a_k a_{k-1} - a_{k-1}^2 = 0
   \]
   Solving this quadratic equation for \( \frac{a_k}{a_{k-1}} \):
   \[
   \left( \frac{a_k}{a_{k-1}} \right)^2 - \left( \frac{a_k}{a_{k-1}} \right) - 1 = 0
   \]
   The positive solution to this equation is:
   \[
   \frac{a_k}{a_{k-1}} = \frac{1 + \sqrt{5}}{2}
   \]
   which is an irrational number. Therefore, \( a_k \) and \( a_{k-1} \) cannot both be integers, contradicting the assumption that \( S \) has more than two elements.

5. Therefore, \( S \) must have exactly two elements. Let \( S = \{a_1, a_2\} \) with \( a_1 < a_2 \). We need:
   \[
   \frac{a_1^2}{a_2 - a_1} \in S
   \]
   Since \( \frac{a_1^2}{a_2 - a_1} \) must be an element of \( S \), it must be either \( a_1 \) or \( a_2 \). If:
   \[
   \frac{a_1^2}{a_2 - a_1} = a_1 \implies a_1^2 = a_1(a_2 - a_1) \implies a_1 = a_2 - a_1 \implies a_2 = 2a_1
   \]

6. Therefore, the only possible sets \( S \) are of the form \( \{n, 2n\} \) for some positive integer \( n \).

The final answer is \( \boxed{ \{n, 2n\} } \) for some positive integer \( n \).