When some number $a^2$ is written in base $b$, the result is $144_b$. $a$ and $b$ also happen to be integer side lengths of a right triangle. If $a$ and $b$ are both less than $20$, find the sum of all possible values of $a$.

1. We start with the given information that \(a^2\) written in base \(b\) is \(144_b\). This means:
   \[
   a^2 = 1 \cdot b^2 + 4 \cdot b + 4
   \]
   Therefore, we have:
   \[
   a^2 = b^2 + 4b + 4
   \]

2. Notice that the right-hand side of the equation can be factored:
   \[
   a^2 = (b + 2)^2
   \]

3. Taking the square root of both sides, we get:
   \[
   a = b + 2
   \]

4. Given that \(a\) and \(b\) are integer side lengths of a right triangle and both are less than 20, we need to find the possible values of \(b\) such that \(a\) is also an integer and less than 20.

5. Since \(a = b + 2\), we need:
   \[
   b + 2 < 20 \implies b < 18
   \]

6. Additionally, since \(a\) and \(b\) are side lengths of a right triangle, we need to ensure that \(a\) and \(b\) satisfy the Pythagorean theorem. However, since \(a = b + 2\), we can directly check the values of \(b\) from 1 to 17.

7. We need to find the sum of all possible values of \(a\):
   \[
   a = b + 2
   \]
   for \(b\) ranging from 1 to 17.

8. The possible values of \(a\) are:
   \[
   a = 3, 4, 5, \ldots, 19
   \]

9. The sum of an arithmetic series can be calculated using the formula:
   \[
   S = \frac{n}{2} (a_1 + a_n)
   \]
   where \(n\) is the number of terms, \(a_1\) is the first term, and \(a_n\) is the last term.

10. Here, \(a_1 = 3\), \(a_n = 19\), and the number of terms \(n = 17\):
    \[
    S = \frac{17}{2} (3 + 19) = \frac{17}{2} \cdot 22 = 17 \cdot 11 = 187
    \]

The final answer is \(\boxed{187}\)