Compute the number of two digit positive integers that are divisible by both of their digits. For example, $36$ is one of these two digit positive integers because it is divisible by both $3$ and $6$.


[i]2021 CCA Math Bonanza Lightning Round #2.4[/i]
1. Let the two-digit number be represented as $\overline{ab}$, where $a$ and $b$ are its digits. This means the number can be written as $10a + b$.
2. For the number $10a + b$ to be divisible by both $a$ and $b$, the following conditions must hold:
   - $a \mid (10a + b)$
   - $b \mid (10a + b)$
3. Since $a \mid (10a + b)$, it implies $a \mid b$ because $10a$ is already divisible by $a$.
4. Since $b \mid (10a + b)$, it implies $b \mid 10a$ because $b \mid b$ is trivially true.
5. Therefore, $b$ must be a divisor of $10a$. Since $a$ is a digit (1 through 9), we need to find the values of $b$ such that $b \mid 10a$ and $a \mid b$.

Let's analyze the possible values of $\frac{b}{a}$:
- $\frac{b}{a} = 1$: This implies $b = a$. The valid pairs $(a, b)$ are $(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (7, 7), (8, 8), (9, 9)$. There are 9 such pairs.
- $\frac{b}{a} = 2$: This implies $b = 2a$. The valid pairs $(a, b)$ are $(1, 2), (2, 4), (3, 6), (4, 8)$. There are 4 such pairs.
- $\frac{b}{a} = 5$: This implies $b = 5a$. The valid pairs $(a, b)$ are $(1, 5), (2, 10)$. However, $b = 10$ is not a valid digit, so the only valid pair is $(1, 5)$. There is 1 such pair.
- $\frac{b}{a} = 10$: This implies $b = 10a$. However, $b = 10a$ is not possible since $b$ must be a single digit (1 through 9).

Summarizing the valid pairs:
- For $\frac{b}{a} = 1$: 9 pairs
- For $\frac{b}{a} = 2$: 4 pairs
- For $\frac{b}{a} = 5$: 1 pair

Adding these, we get a total of $9 + 4 + 1 = 14$ valid pairs.

The final answer is $\boxed{14}$.