The remainder when we divide $a \sqrt{2}+b \sqrt{3}$ by $\sqrt{2}$ is defined to be the smallest non-negative real $r_{1}$ such that $\frac{a \sqrt{2}+b \sqrt{3}-r_{1}}{\sqrt{2}}$ is integral. As $\frac{x}{\sqrt{2}}$ is integral iff $x$ is an integral multiple of $\sqrt{2}$, it follows that $r_{1}=b \sqrt{3}-c \sqrt{2}$, for some integer $c$. Furthermore given any real $r$ such that $\frac{a \sqrt{2}+b \sqrt{3}-r}{\sqrt{2}}$ is integral, we may add or subtract $\sqrt{2}$ to $r$ and the fraction remains an integer. Thus, the smallest non-negative real $r_{1}$ such that the fraction is an integer must satisfy $0 \leq r_{1}<\sqrt{2}$. Similarly, we find $r_{2}=a \sqrt{2}-d \sqrt{3}$ for some integer $d$ and $0 \leq r_{2}<\sqrt{3}$. Since $r_{1}+r_{2}=\sqrt{2}$, then $$(a-c) \sqrt{2}+(b-d) \sqrt{3}=\sqrt{2} \Longleftrightarrow a-c=1 \text { and } b-d=0$$ Finally, substituting in $c=a-1$ and $d=b$ plugging back into our bounds for $r_{1}$ and $r_{2}$, we get $$\left\{\begin{array}{l} 0 \leq b \sqrt{3}-(a-1) \sqrt{2}<\sqrt{2} \\ 0 \leq a \sqrt{2}-b \sqrt{3}<\sqrt{3} \end{array}\right.$$ or $$\left\{\begin{array}{l} (a-1) \sqrt{2} \leq b \sqrt{3} \\ b \sqrt{3}<a \sqrt{2} \\ b \sqrt{3} \leq a \sqrt{2} \\ a \sqrt{2}<(b+1) \sqrt{3} \end{array}\right.$$ Note that $b \sqrt{3}<a \sqrt{2} \Longrightarrow b \sqrt{3} \leq a \sqrt{2}$ and $$(a-1) \sqrt{2} \leq b \sqrt{3} \Longrightarrow a \sqrt{2} \leq b \sqrt{3}+\sqrt{2}<b \sqrt{3}+\sqrt{3}=(b+1) \sqrt{3}$$ so the last two inequalities are redundant. We are left with $$(a-1) \sqrt{2} \leq b \sqrt{3}<a \sqrt{2}$$ Since the non-negative number line is partitioned by intervals of the form $[(a-1) \sqrt{2}, a \sqrt{2})$ for positive integers $a$, for any positive integer $b$, we can find a positive integer $a$ that satisfies the inequalities. As clearly $a>b$, it remains to find the number of $b$ such that $a \leq 20$. This is bounded by $$b \sqrt{3}<a \sqrt{2} \leq 20 \sqrt{2} \Longleftrightarrow b<\frac{20 \sqrt{2}}{\sqrt{3}} \Longrightarrow b \leq 16$$ so there are 16 values of $b$ and thus 16 ordered pairs of positive integers $(a, b)$ that satisfy the problem.