Find all pairs of positive integers $(a, b)$ such that $$11ab \le a^3 - b^3 \le 12ab.$$
To solve the problem, we need to find all pairs of positive integers \((a, b)\) such that:
\[ 11ab \le a^3 - b^3 \le 12ab. \]

1. **Rewriting the inequalities:**
   We start by rewriting the given inequalities:
   \[ 11ab \le a^3 - b^3 \le 12ab. \]
   We can factorize \(a^3 - b^3\) using the difference of cubes:
   \[ a^3 - b^3 = (a - b)(a^2 + ab + b^2). \]
   Substituting this into the inequalities, we get:
   \[ 11ab \le (a - b)(a^2 + ab + b^2) \le 12ab. \]

2. **Analyzing the left inequality:**
   \[ 11ab \le (a - b)(a^2 + ab + b^2). \]
   Since \(a\) and \(b\) are positive integers, we can divide both sides by \(ab\) (assuming \(b \neq 0\)):
   \[ 11 \le \frac{(a - b)(a^2 + ab + b^2)}{ab}. \]
   Simplifying the right-hand side:
   \[ 11 \le \frac{a^3 - b^3}{ab} = \frac{a^3}{ab} - \frac{b^3}{ab} = \frac{a^2}{b} - b. \]
   Therefore, we have:
   \[ 11 \le \frac{a^2}{b} - b. \]

3. **Analyzing the right inequality:**
   \[ (a - b)(a^2 + ab + b^2) \le 12ab. \]
   Similarly, we divide both sides by \(ab\):
   \[ \frac{(a - b)(a^2 + ab + b^2)}{ab} \le 12. \]
   Simplifying the left-hand side:
   \[ \frac{a^3 - b^3}{ab} = \frac{a^3}{ab} - \frac{b^3}{ab} = \frac{a^2}{b} - b. \]
   Therefore, we have:
   \[ \frac{a^2}{b} - b \le 12. \]

4. **Combining the inequalities:**
   Combining the results from steps 2 and 3, we get:
   \[ 11 \le \frac{a^2}{b} - b \le 12. \]
   Let \(k = \frac{a^2}{b} - b\). Then:
   \[ 11 \le k \le 12. \]
   This implies that \(k\) can be either 11 or 12.

5. **Case analysis:**
   - **Case 1: \(k = 11\)**
     \[ \frac{a^2}{b} - b = 11. \]
     \[ \frac{a^2}{b} = 11 + b. \]
     \[ a^2 = b(11 + b). \]
     Since \(a\) and \(b\) are positive integers, \(b(11 + b)\) must be a perfect square. Let's test small values of \(b\):
     - For \(b = 2\):
       \[ a^2 = 2(11 + 2) = 2 \cdot 13 = 26, \]
       which is not a perfect square.
     - For \(b = 3\):
       \[ a^2 = 3(11 + 3) = 3 \cdot 14 = 42, \]
       which is not a perfect square.
     - For \(b = 4\):
       \[ a^2 = 4(11 + 4) = 4 \cdot 15 = 60, \]
       which is not a perfect square.
     - For \(b = 5\):
       \[ a^2 = 5(11 + 5) = 5 \cdot 16 = 80, \]
       which is not a perfect square.
     - For \(b = 1\):
       \[ a^2 = 1(11 + 1) = 1 \cdot 12 = 12, \]
       which is not a perfect square.
     - For \(b = 2\):
       \[ a^2 = 2(11 + 2) = 2 \cdot 13 = 26, \]
       which is not a perfect square.
     - For \(b = 3\):
       \[ a^2 = 3(11 + 3) = 3 \cdot 14 = 42, \]
       which is not a perfect square.
     - For \(b = 4\):
       \[ a^2 = 4(11 + 4) = 4 \cdot 15 = 60, \]
       which is not a perfect square.
     - For \(b = 5\):
       \[ a^2 = 5(11 + 5) = 5 \cdot 16 = 80, \]
       which is not a perfect square.
     - For \(b = 6\):
       \[ a^2 = 6(11 + 6) = 6 \cdot 17 = 102, \]
       which is not a perfect square.
     - For \(b = 7\):
       \[ a^2 = 7(11 + 7) = 7 \cdot 18 = 126, \]
       which is not a perfect square.
     - For \(b = 8\):
       \[ a^2 = 8(11 + 8) = 8 \cdot 19 = 152, \]
       which is not a perfect square.
     - For \(b = 9\):
       \[ a^2 = 9(11 + 9) = 9 \cdot 20 = 180, \]
       which is not a perfect square.
     - For \(b = 10\):
       \[ a^2 = 10(11 + 10) = 10 \cdot 21 = 210, \]
       which is not a perfect square.
     - For \(b = 11\):
       \[ a^2 = 11(11 + 11) = 11 \cdot 22 = 242, \]
       which is not a perfect square.
     - For \(b = 12\):
       \[ a^2 = 12(11 + 12) = 12 \cdot 23 = 276, \]
       which is not a perfect square.
     - For \(b = 13\):
       \[ a^2 = 13(11 + 13) = 13 \cdot 24 = 312, \]
       which is not a perfect square.
     - For \(b = 14\):
       \[ a^2 = 14(11 + 14) = 14 \cdot 25 = 350, \]
       which is not a perfect square.
     - For \(b = 15\):
       \[ a^2 = 15(11 + 15) = 15 \cdot 26 = 390, \]
       which is not a perfect square.
     - For \(b = 16\):
       \[ a^2 = 16(11 + 16) = 16 \cdot 27 = 432, \]
       which is not a perfect square.
     - For \(b = 17\):
       \[ a^2 = 17(11 + 17) = 17 \cdot 28 = 476, \]
       which is not a perfect square.
     - For \(b = 18\):
       \[ a^2 = 18(11 + 18) = 18 \cdot 29 = 522, \]
       which is not a perfect square.
     - For \(b = 19\):
       \[ a^2 = 19(11 + 19) = 19 \cdot 30 = 570, \]
       which is not a perfect square.
     - For \(b = 20\):
       \[ a^2 = 20(11 + 20) = 20 \cdot 31 = 620, \]
       which is not a perfect square.
     - For \(b = 21\):
       \[ a^2 = 21(11 + 21) = 21 \cdot 32 = 672, \]
       which is not a perfect square.
     - For \(b = 22\):
       \[ a^2 = 22(11 + 22) = 22 \cdot 33 = 726, \]
       which is not a perfect square.
     - For \(b = 23\):
       \[ a^2 = 23(11 + 23) = 23 \cdot 34 = 782, \]
       which is not a perfect square.
     - For \(b = 24\):
       \[ a^2 = 24(11 + 24) = 24 \cdot 35 = 840, \]
       which is not a perfect square.
     - For \(b = 25\):
       \[ a^2 = 25(11 + 25) = 25 \cdot 36 = 900, \]
       which is a perfect square. Therefore, \(a = 30\).

     So, one solution is \((a, b) = (30, 25)\).

   - **Case 2: \(k = 12\)**
     \[ \frac{a^2}{b} - b = 12. \]
     \[ \frac{a^2}{b} = 12 + b. \]
     \[ a^2 = b(12 + b). \]
     Since \(a\) and \(b\) are positive integers, \(b(12 + b)\) must be a perfect square. Let's test small values of \(b\):
     - For \(b = 1\):
       \[ a^2 = 1(12 + 1) = 1 \cdot 13 = 13, \]
       which is not a perfect square.
     - For \(b = 2\):
       \[ a^2 = 2(12 + 2) = 2 \cdot 14 = 28, \]
       which is not a perfect square.
     - For \(b = 3\):
       \[ a^2 = 3(12 + 3) = 3 \cdot 15 = 45, \]
       which is not a perfect square.
     - For \(b = 4\):
       \[ a^2 = 4(12 + 4) = 4 \cdot 16 = 64, \]
       which is a perfect square. Therefore, \(a = 8\).

     So, another solution is \((a, b) = (8, 4)\).

The final answer is \( \boxed{ (a, b) = (30, 25) } \) and \((a, b) = (8, 4)\).