Find all finite sets of positive integers with at least two elements such that for any two numbers $ a$, $ b$ ($ a > b$) belonging to the set, the number $ \frac {b^2}{a \minus{} b}$ belongs to the set, too.
1. Let \( S \) be a finite set of positive integers with at least two elements that satisfies the given condition: for any two numbers \( a \) and \( b \) (with \( a > b \)) in \( S \), the number \( \frac{b^2}{a - b} \) also belongs to \( S \).

2. Select \( x \) to be the minimal element in \( S \). Consider another element \( a \in S \) such that \( a > x \).

3. By the minimality of \( x \), we have:
   \[
   \frac{x^2}{a - x} \geq x
   \]
   This implies:
   \[
   x^2 \geq x(a - x) \implies x \geq a - x \implies a \leq 2x
   \]

4. Suppose for the sake of contradiction that there exists an \( a \) such that \( x < a < 2x \). Then, the condition implies:
   \[
   \frac{x^2}{a - x} \leq 2x
   \]
   This implies:
   \[
   x^2 \leq 2x(a - x) \implies x \leq 2a - 2x \implies a \geq \frac{3x}{2}
   \]

5. Therefore, for all \( a \in S \), we have:
   \[
   \frac{3x}{2} \leq a \leq 2x
   \]

6. Similarly, because:
   \[
   \frac{x^2}{a - x} \geq \frac{3x}{2}
   \]
   and for all \( a \neq 2x \), we must have:
   \[
   a \leq \frac{5x}{3}
   \]

7. Continuing in this fashion, we may obtain:
   \[
   \frac{F_{2n} x}{F_{2n-1}} \leq a \leq \frac{F_{2n-1} x}{F_{2n-2}}
   \]
   for all positive integers \( n \), where \( F_n \) denotes the \( n \)-th Fibonacci number.

8. But:
   \[
   \lim_{n \to \infty} \frac{F_n x}{F_{n-1} x} = \phi
   \]
   where \( \phi \) is the golden ratio, which is irrational. Therefore, such a positive integer \( a \) cannot exist given a positive integer \( x \).

9. The only possible set \( S \) that satisfies the given conditions is \( \{x, 2x\} \) for any positive integer \( x \).

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