Denote by $A$ and $B$ the set of numbers written on the blackboard $A$ and $B$, respectively. Denote also by $|X|$ the number of elements belonging to the set $X$. If we let
$$A = \{2, 3, 4, 5, 6, 8, 9, 10, 12, 15, 16, 18, 20\}, \quad B = \{7, 11, 13, 17, 19\},$$
then we see that arbitrary pair of numbers $(a, b)$ with $a \in A$ and $b \in B$ are relatively prime, and that we have $|A||B| = 65$. We will show in the sequel that $65$ is the maximum possible value for the product $|A||B|$.

So, suppose that for some choice of $A$ and $B$, $|A||B| \ge 66$, and we will show that this will lead to a contradiction. From $|A| + |B| \ge 2\sqrt{|A||B|} \ge 2\sqrt{66}$, it follows that we must have $|A| + |B| \ge 17$. Therefore, we conclude that there are at most $2$ numbers among the integers, greater than or equal to $2$ and less than or equal to $20$, which belong neither to $A$ nor to $B$. Hence, at least one number among $6, 12, 18$ must belong either to $A$ or to $B$. We may suppose without loss of generality that the former is the case. By the assumption of the problem, multiples of a same positive number cannot belong to both $A$ and $B$. Since at least one of the numbers $6, 12, 18$ belongs to $A$, we conclude that all the multiples of $2$ and all the multiples of $3$ must also belong to $A$. Since there are only six numbers $5, 7, 11, 13, 17, 19$, among the numbers greater than or equal to $2$ and less than or equal to $20$, which are neither multiples of $2$ nor multiples of $3$, we conclude that $|B| \le 6$ must be satisfied. If we assume that $|B| \le 4$, then we get $|A||B| \le (19-|B|)|B| \le 60$, which contradicts our assumption. Therefore, we must have either $|B| = 5$ or $|B| = 6$.

* When $|B| = 5$:

In this case, we must have either $5$ or $7$ in the set $B$. But then at least one of the numbers $10$ or $14$ cannot be in the set $A$, so that we have $|A| \le 13$, which implies that $|A||B| \le 65$, which contradicts our assumption.

* When $|B| = 6$:

In this case, we have $B = \{5, 7, 11, 13, 17, 19\}$, so neither $10$ nor $14$ can belong to the set $A$, which means that $|A| \le 9$, and this also leads to the contradiction to our assumption.

Thus we conclude that $65$ is the maximum possible value that the product $|A||B|$ can take.