Let $S$ be the set of positive integer divisors of $20^9.$ Three numbers are chosen independently and at random from the set $S$ and labeled $a_1,a_2,$ and $a_3$ in the order they are chosen. The probability that both $a_1$ divides $a_2$ and $a_2$ divides $a_3$ is $\frac mn,$ where $m$ and $n$ are relatively prime positive integers. Find $m.$
1. **Identify the set \( S \)**:
   The set \( S \) consists of all positive integer divisors of \( 20^9 \). Since \( 20 = 2^2 \cdot 5 \), we have:
   \[
   20^9 = (2^2 \cdot 5)^9 = 2^{18} \cdot 5^9
   \]
   Therefore, the divisors of \( 20^9 \) are of the form \( 2^m \cdot 5^n \) where \( 0 \leq m \leq 18 \) and \( 0 \leq n \leq 9 \).

2. **Define the chosen numbers**:
   Let \( a_1 = 2^{m_1} \cdot 5^{n_1} \), \( a_2 = 2^{m_2} \cdot 5^{n_2} \), and \( a_3 = 2^{m_3} \cdot 5^{n_3} \). We need to find the probability that \( a_1 \) divides \( a_2 \) and \( a_2 \) divides \( a_3 \). This translates to:
   \[
   m_1 \leq m_2 \leq m_3 \quad \text{and} \quad n_1 \leq n_2 \leq n_3
   \]

3. **Calculate the probability for \( m \)**:
   - **All \( m_i \) are distinct**:
     The number of ways to choose 3 distinct \( m_i \) from 19 possible values (0 to 18) is \( \binom{19}{3} \). Each choice can be ordered in exactly one way to satisfy \( m_1 \leq m_2 \leq m_3 \). Thus, the probability is:
     \[
     \frac{\binom{19}{3}}{19^3} = \frac{\frac{19 \cdot 18 \cdot 17}{6}}{19^3} = \frac{19 \cdot 18 \cdot 17}{6 \cdot 19^3} = \frac{18 \cdot 17}{6 \cdot 19^2} = \frac{51}{19^2}
     \]

   - **Exactly two \( m_i \) are equal**:
     - If \( m_1 = m_2 \neq m_3 \):
       For each \( k \) from 0 to 17, there are \( 18 - k \) choices for \( m_3 \). Summing these gives:
       \[
       \sum_{k=0}^{17} (18 - k) = 18 + 17 + \cdots + 1 = \frac{18 \cdot 19}{2} = 171
       \]
     - If \( m_1 \neq m_2 = m_3 \):
       For each \( k \) from 1 to 18, there are \( k \) choices for \( m_1 \). Summing these gives:
       \[
       \sum_{k=1}^{18} k = 1 + 2 + \cdots + 18 = \frac{18 \cdot 19}{2} = 171
       \]
     The total number of ways for exactly two \( m_i \) to be equal is \( 171 + 171 = 342 \). The probability is:
     \[
     \frac{342}{19^3} = \frac{342}{19^3} = \frac{18}{19^2}
     \]

   - **All three \( m_i \) are equal**:
     There are 19 ways to choose the value, so the probability is:
     \[
     \frac{19}{19^3} = \frac{1}{19^2}
     \]

   - **Total probability for \( m \)**:
     \[
     \frac{51}{19^2} + \frac{18}{19^2} + \frac{1}{19^2} = \frac{70}{19^2}
     \]

4. **Calculate the probability for \( n \)**:
   The same logic applies to \( n \) with the range from 0 to 9:
   - **All \( n_i \) are distinct**:
     \[
     \frac{\binom{10}{3}}{10^3} = \frac{\frac{10 \cdot 9 \cdot 8}{6}}{10^3} = \frac{10 \cdot 9 \cdot 8}{6 \cdot 10^3} = \frac{12}{100} = \frac{3}{25}
     \]

   - **Exactly two \( n_i \) are equal**:
     - If \( n_1 = n_2 \neq n_3 \):
       \[
       \sum_{k=0}^{8} (9 - k) = 9 + 8 + \cdots + 1 = \frac{9 \cdot 10}{2} = 45
       \]
     - If \( n_1 \neq n_2 = n_3 \):
       \[
       \sum_{k=1}^{9} k = 1 + 2 + \cdots + 9 = \frac{9 \cdot 10}{2} = 45
       \]
     The total number of ways for exactly two \( n_i \) to be equal is \( 45 + 45 = 90 \). The probability is:
     \[
     \frac{90}{10^3} = \frac{9}{100}
     \]

   - **All three \( n_i \) are equal**:
     There are 10 ways to choose the value, so the probability is:
     \[
     \frac{10}{10^3} = \frac{1}{100}
     \]

   - **Total probability for \( n \)**:
     \[
     \frac{3}{25} + \frac{9}{100} + \frac{1}{100} = \frac{12}{100} = \frac{3}{25}
     \]

5. **Combine the probabilities**:
   The combined probability is:
   \[
   \frac{70}{19^2} \cdot \frac{3}{25} = \frac{210}{25 \cdot 19^2} = \frac{210}{4750} = \frac{21}{475} = \frac{77}{5 \cdot 19^2}
   \]

The final answer is \( \boxed{77} \)