Find all pairs $(a, b)$ of positive integers such that $a \le b$ and
$$ \gcd(x, a) \gcd(x, b) = \gcd(x, 20) \gcd(x, 22) $$ 
holds for every positive integer $x$.
1. **Claim 1: If \( p \mid ab \) is a prime, then \( p \in \{2, 5, 11\} \).**

   To prove this, assume \( q \mid ab \) with \( q \notin \{2, 5, 11\} \). Consider \( x = q \). Then:
   \[
   \gcd(q, a) \gcd(q, b) = \gcd(q, 20) \gcd(q, 22)
   \]
   Since \( q \notin \{2, 5, 11\} \), we have \( \gcd(q, 20) = 1 \) and \( \gcd(q, 22) = 1 \). Thus, the right-hand side (RHS) is 1. However, since \( q \mid ab \), either \( q \mid a \) or \( q \mid b \), making the left-hand side (LHS) divisible by \( q \), which is a contradiction. Therefore, \( p \in \{2, 5, 11\} \).

2. **Claim 2: \( ab = 440 \).**

   We need to show that \( v_p(ab) = v_p(440) \) for \( p \in \{2, 5, 11\} \).

   - For \( p = 5 \) (also applies to \( p = 11 \)):
     - If \( v_5(ab) = 0 \), then \( 5 \nmid \text{LHS} \).
     - If \( v_5(ab) \geq 2 \), taking \( x = 125 \), we see that \( 25 \mid \text{LHS} \) but \( 25 \nmid \text{RHS} \), which is a contradiction.

   - For \( p = 2 \):
     - If \( v_2(ab) \leq 2 \), taking \( x = 8 \), we get a contradiction since \( v_2(\text{LHS}) \neq v_2(\text{RHS}) \).
     - If \( v_2(ab) \geq 4 \), taking \( x = 2^k \) for some large \( k \), we get a contradiction.

   Hence, \( v_2(ab) = 3 \), \( v_5(ab) = 1 \), and \( v_11(ab) = 1 \), so \( ab = 2^3 \cdot 5 \cdot 11 = 440 \).

3. **Show \( a \) and \( b \) are simultaneously even.**

   Assume \( 2 \nmid a \). Taking \( x = 4 \), we get \( v_2(\text{LHS}) = 2 \) whereas \( v_2(\text{RHS}) = 3 \), which is not possible. Therefore, both \( a \) and \( b \) must be even.

4. **Possible pairs \((a, b)\):**

   The pairs \((a, b)\) that satisfy \( ab = 440 \) and both \( a \) and \( b \) are even are:
   \[
   (2, 220), (4, 110), (10, 44), (20, 22)
   \]

   Verification:
   - For \((2, 220)\):
     \[
     \gcd(x, 2) \gcd(x, 220) = \gcd(x, 20) \gcd(x, 22)
     \]
     This holds for all \( x \).

   - For \((4, 110)\):
     \[
     \gcd(x, 4) \gcd(x, 110) = \gcd(x, 20) \gcd(x, 22)
     \]
     This holds for all \( x \).

   - For \((10, 44)\):
     \[
     \gcd(x, 10) \gcd(x, 44) = \gcd(x, 20) \gcd(x, 22)
     \]
     This holds for all \( x \).

   - For \((20, 22)\):
     \[
     \gcd(x, 20) \gcd(x, 22) = \gcd(x, 20) \gcd(x, 22)
     \]
     This holds for all \( x \).

The final answer is \( \boxed{ (2, 220), (4, 110), (10, 44), (20, 22) } \).