1. Define the remainders $r_1$ and $r_2$ as follows:
$$r_1 = a\sqrt{2} + b\sqrt{3} - k_1\sqrt{2}$$
where $k_1$ is an integer such that $0 \leq a\sqrt{2} + b\sqrt{3} - k_1\sqrt{2} < \sqrt{2}$. Simplifying, we get:
$$r_1 = b\sqrt{3} - (k_1 - a)\sqrt{2}$$
Similarly, for $r_2$:
$$r_2 = a\sqrt{2} + b\sqrt{3} - k_2\sqrt{3}$$
where $k_2$ is an integer such that $0 \leq a\sqrt{2} + b\sqrt{3} - k_2\sqrt{3} < \sqrt{3}$. Simplifying, we get:
$$r_2 = a\sqrt{2} - (k_2 - b)\sqrt{3}$$

2. Given that $r_1 + r_2 = \sqrt{2}$, we substitute the expressions for $r_1$ and $r_2$:
$$b\sqrt{3} - (k_1 - a)\sqrt{2} + a\sqrt{2} - (k_2 - b)\sqrt{3} = \sqrt{2}$$
Simplifying, we get:
$$b\sqrt{3} - k_1\sqrt{2} + a\sqrt{2} - k_2\sqrt{3} + a\sqrt{2} + b\sqrt{3} = \sqrt{2}$$
$$2b\sqrt{3} + 2a\sqrt{2} - k_1\sqrt{2} - k_2\sqrt{3} = \sqrt{2}$$

3. To satisfy the condition $r_1 + r_2 = \sqrt{2}$, we need:
$$b\sqrt{3} - (a-1)\sqrt{2} = b\sqrt{3} - a\sqrt{2} + \sqrt{2} < \sqrt{2}$$
and
$$0 \leq a\sqrt{2} - b\sqrt{3} < \sqrt{3}$$

4. Combining these conditions, we get:
$$0 < a\sqrt{2} - b\sqrt{3} \leq \sqrt{2}$$

5. For each $b$, there exists exactly one value of $a$ such that the inequality holds. We need to find the number of $b$ such that there exists an $a \leq 20$.

6. We need to check the range of $b$ for which $a \leq 20$. Since $a$ and $b$ are positive integers, we need to find the number of $b$ such that:
$$0 < a\sqrt{2} - b\sqrt{3} \leq \sqrt{2}$$
This implies:
$$b\sqrt{3} < a\sqrt{2} \leq b\sqrt{3} + \sqrt{2}$$

7. For $a \leq 20$, we need to find the number of $b$ such that:
$$b\sqrt{3} < 20\sqrt{2}$$
Simplifying, we get:
$$b < \frac{20\sqrt{2}}{\sqrt{3}} \approx 16.33$$

8. Therefore, $b$ can take integer values from 1 to 16. For each $b$, there is exactly one $a$ such that $a \leq 20$.

The final answer is $\boxed{16}$.