Let $r$ be a positive integer. Find the smallest positive integer $m$ satisfying the condition: For all sets $A_1, A_2, \dots, A_r$ with $A_i \cap A_j = \emptyset$, for all $i \neq j$, and $\bigcup_{k = 1}^{r} A_k = \{ 1, 2, \dots, m \}$, there exists $a, b \in A_k$ for some $k$ such that $1 \leq \frac{b}{a} \leq 1 + \frac{1}{2022}$.
To find the smallest positive integer \( m \) such that for any partition of the set \(\{1, 2, \dots, m\}\) into \( r \) disjoint subsets \( A_1, A_2, \dots, A_r \), there exist \( a, b \in A_k \) for some \( k \) such that \( 1 \leq \frac{b}{a} \leq 1 + \frac{1}{2022} \), we proceed as follows:

1. **Establish the condition for \( m = 2023r \):**
   - Consider the set \(\{1, 2, \dots, 2023r\}\).
   - Partition this set into \( r \) subsets \( A_1, A_2, \dots, A_r \) such that each subset contains consecutive integers. For example, \( A_1 = \{1, 2, \dots, 2023\} \), \( A_2 = \{2024, 2025, \dots, 4046\} \), and so on.
   - Within each subset \( A_k \), consider two elements \( a = 2022r + x \) and \( b = 2022r + y \) where \( 0 \leq x < y \leq r \).

2. **Verify the ratio condition:**
   - We need to show that \( 1 \leq \frac{b}{a} \leq 1 + \frac{1}{2022} \).
   - Calculate the ratio:
     \[
     \frac{b}{a} = \frac{2022r + y}{2022r + x}
     \]
   - Since \( 0 \leq x < y \leq r \), we have:
     \[
     \frac{2022r + y}{2022r + x} \leq 1 + \frac{y - x}{2022r + x}
     \]
   - Given \( y - x \leq r \) and \( 2022r + x \geq 2022r \), we get:
     \[
     \frac{y - x}{2022r + x} \leq \frac{r}{2022r} = \frac{1}{2022}
     \]
   - Therefore:
     \[
     \frac{2022r + y}{2022r + x} \leq 1 + \frac{1}{2022}
     \]

3. **Consider \( m < 2023r \):**
   - Suppose \( m < 2023r \) and the statement is still true.
   - Consider sets \( A_i = \{i \pmod{r}\} \). For any such set, the difference \( b - a \geq r \).
   - Hence:
     \[
     \frac{1}{2022} \geq \frac{b - a}{a} \geq \frac{r}{a}
     \]
   - This implies:
     \[
     m \geq b \geq a + r \geq 2023r
     \]
   - This yields a contradiction since \( m < 2023r \).

Thus, the smallest \( m \) that satisfies the condition is \( 2023r \).

The final answer is \(\boxed{2023r}\).