Question 239, Let $M$ be a set composed of a finite number of positive integers, and $M=U_{i=1}^{20} A_{i}=U_{i=1}^{20} B_{i}$, where $A_{i} \neq \emptyset$, $B_{i} \neq \emptyset, i=1, 2, \ldots, 20$, and for any $1 \leq i<j \leq 20$, we have $A_{i} \cap A_{j}=\emptyset, B_{i} \cap B_{j}=\emptyset$.

It is known that for any $1 \leq i \leq 20,1 \leq j \leq 20$, if $A_{i} \cap B_{j}=\emptyset$, then $\left|A_{i} \cup B_{j}\right| \geq 18$, find the minimum number of elements in the set $M$.

Question 239, Solution: Let's assume among $A_{1}, A_{2}, \ldots, A_{20}, B_{1}, B_{2}, \ldots, B_{20}$, the number of elements in the set with the fewest elements is $k$, and let's assume $\left|A_{1}\right|=k$.

Let $B_{1}, B_{2}, \ldots, B_{20}$ contain exactly $t$ subsets that have an empty intersection with $A_{1}$. Noting that $\left|A_{1}\right|=k$, $A_{1}$ can have a non-empty intersection with at most $k$ of the subsets $B_{1}, B_{2}, \ldots, B_{20}$. Therefore, $t \geq 20-k$. Without loss of generality, assume $B_{1}, B_{2}, \ldots, B_{t}$ all have an empty intersection with $A_{1}$. Then $\left|B_{i}\right| \geq 18-k (1 \leq i \leq t)$, and $\left|B_{i}\right| \geq k (1 \leq i \leq 20)$. Thus, we have:
$$
|M|=\sum_{i=1}^{20}\left|B_{i}\right| \geq t(18-k)+(20-t) k=180-2(k-9)(t-10)
$$

And we also have:
$$
|M|=\sum_{i=1}^{20}\left|B_{i}\right| \geq 20 k
$$

If $k \geq 9$, then according to (2), $|M| \geq 180$;
If $k < 9$, then according to (1), we have:
$$
\begin{array}{l}
|M| \geq 180-2(k-9)(t-10) \\
=180+2(9-k)(t-10) \\
\geq 180+2(9-k)[(20-k)-10] \\
=180+2(9-k)(10-k) \geq 180
\end{array}
$$

Therefore, we always have $|M| \geq 180$.
On the other hand, when $|M|=180$, arbitrarily divide $M$ into 20 subsets, each with 9 elements, denoted as $A_{1}, A_{2}, \ldots, A_{20}$; then arbitrarily divide $M$ into 20 subsets, each with 9 elements, denoted as $B_{1}, B_{2}, \ldots, B_{20}$. Clearly, this satisfies the conditions of the problem. In summary, the minimum value of $|M|$ is 180.