Example 3 There are two blackboards $A$ and $B$, on which are written several different integers no less than 2 and no greater than 20. If each time a number is taken from blackboard $A$ and a number from blackboard $B$, these two numbers are definitely coprime. Find the possible maximum value of the product of the number of integers written on blackboard $A$ and the number of integers written on blackboard $B$.
Let the set of integers written on blackboard $A$ be $A$, and the set of integers written on blackboard $B$ be $B$. Use $|X|$ to denote the number of elements in set $X$.
$$
\begin{array}{l}
\text { When } A=\{2,3,4,5,6,8,9,10,12,15,16,18,20\}, \\
B=\{7,11,13,17,19\}
\end{array}
$$

the elements of set $A$ are coprime with the elements of set $B$, so sets $A$ and $B$ meet the requirements, and $|A||B|=65$.
Now, we prove that this is the maximum.
Assume $|A||B| \geqslant 66$. Then
$$
\begin{array}{l}
|A|+|B| \geqslant 2 \sqrt{|A||B|} \geqslant 2 \sqrt{66} \\
\Rightarrow|A|+|B| \geqslant 17 .
\end{array}
$$

Thus, among the integers not greater than 20 and not less than 2, the number of integers that are neither in set $A$ nor in set $B$ is no more than 2. Therefore, at least one of $6, 12, 18$ belongs to set $A$ or set $B$. Without loss of generality, assume it belongs to set $A$.

Since any element in set $A$ is coprime with any element in set $B$, multiples of 2 can only belong to set $A$ or set $B$, and multiples of 3 can only belong to set $A$ or set $B$.

Since one of $6, 12, 18$ belongs to set $A$, multiples of 2 and multiples of 3 must belong to set $A$.

Among the integers not less than 2 and not greater than 20, the numbers that are neither multiples of 2 nor multiples of 3 are only 6, namely $5, 7, 11, 13, 17, 19$, so $|B| \leqslant 6$.
If $|B| \leqslant 4$, then
$$
|A||B| \leqslant(19-|B|)|B| \leqslant 15 \times 4=60,
$$
which is a contradiction.
Thus, $|B|=5$ or 6.
When $|B|=5$, since five of $5, 6, 7, 11, 13, 17, 19$ belong to $B$, 5 or 7 belongs to set $B$.
Therefore, 10 or 14 does not belong to set $A$.
Then $|A| \leqslant|B|$.
Thus, $|A||B| \leqslant 9 \times 6<66$, which is a contradiction.
In conclusion, the assumption is false.
Therefore, the maximum value that $|A||B|$ can reach is 65.
[Comment] This problem essentially involves distributing prime factors to sets $A$ and $B$. To determine the maximum value, the condition $|A|+|B| \geqslant 17$ is added, and the process of deriving a contradiction also confirms the existence of the maximum value when $A$ and $B$ are determined.