Let $a$ and $b$ be positive integers. As is known, the division of of $a \cdot  b$ with $a + b$ determines  integers $q$ and $r$ uniquely such that $a \cdot b = q (a + b) + r$ and $0 \le  r <a + b$. Find all pairs $(a, b)$ for which $q^2 + r = 2011$.
1. We start with the given equations and inequalities:
   \[
   ab = (a + b)q + r
   \]
   \[
   q^2 + r = 2011
   \]
   \[
   0 \leq r < a + b
   \]

2. We assume without loss of generality that \(a \leq b\). This assumption is valid because \(a\) and \(b\) are symmetric in the given equations and inequalities.

3. From the first equation, we can rewrite it as:
   \[
   ab = (a + b)q + r
   \]
   Rearranging, we get:
   \[
   ab - aq - bq = r
   \]
   \[
   (a - q)(b - q) = q^2 + r
   \]

4. Using the second equation \(q^2 + r = 2011\), we substitute \(q^2 + r\) in the above equation:
   \[
   (a - q)(b - q) = 2011
   \]

5. Since 2011 is a prime number, the possible pairs \((a - q, b - q)\) are \((1, 2011)\) and \((-1, -2011)\).

6. If \((a - q, b - q) = (-1, -2011)\), then:
   \[
   b = q - 2011
   \]
   Since \(q < 45\) (as derived from \(q^2 + r = 2011\) and \(0 \leq r < a + b\)), we have:
   \[
   b = q - 2011 < 45 - 2011 < 0
   \]
   This is a contradiction because \(b\) must be a positive integer. Therefore, \((a - q, b - q) = (-1, -2011)\) is not possible.

7. Hence, the only valid pair is \((a - q, b - q) = (1, 2011)\). This gives us:
   \[
   a = q + 1
   \]
   \[
   b = q + 2011
   \]

8. Substituting \(a = q + 1\) and \(b = q + 2011\) into the first equation:
   \[
   r = ab - (a + b)q
   \]
   \[
   r = (q + 1)(q + 2011) - [(q + 1) + (q + 2011)]q
   \]
   \[
   r = q^2 + 2012q + 2011 - (2q^2 + 2012q)
   \]
   \[
   r = 2011 - q^2
   \]

9. This value of \(r\) satisfies the second equation \(q^2 + r = 2011\). Therefore, the system of equations has the solutions:
   \[
   (a, b, q, r) = (s + 1, s + 2011, s, 2011 - s^2)
   \]
   where \(s\) is an integer.

10. To satisfy the inequality \(0 \leq r < a + b\), we need:
    \[
    0 \leq 2011 - s^2 < 2s + 2012
    \]
    This is satisfied if and only if \(0 \leq s \leq 44\).

11. By setting \(t = s + 1\), the pairs \((a, b)\) satisfying the conditions are:
    \[
    (a, b) = (t, t + 2012) \quad \text{or} \quad (t + 2012, t) \quad \text{for} \quad 1 \leq t \leq 45
    \]

The final answer is \( \boxed{ (a, b) = (t, t + 2012) \quad \text{or} \quad (t + 2012, t) \quad \text{for} \quad 1 \leq t \leq 45 } \).