15. Let $M$ be a set composed of a finite number of positive integers
$$
\text { such that, } \begin{aligned}
M & =\bigcup_{i=1}^{20} A_{i}=\bigcup_{i=1}^{20} B_{i}, \text { where, } \\
A_{i} & \neq \varnothing, B_{i} \neq \varnothing(i=1,2, \cdots, 20),
\end{aligned}
$$

and satisfies:
(1) For any $1 \leqslant i<j \leqslant 20$,
$$
A_{i} \cap A_{j}=\varnothing, B_{i} \cap B_{j}=\varnothing \text {; }
$$
(2) For any $1 \leqslant i \leqslant 20,1 \leqslant j \leqslant 20$, if $A_{i} \cap B_{j}=\varnothing$, then $\left|A_{i} \cup B_{j}\right| \geqslant 18$.

Find the minimum number of elements in the set $M$ ($|X|$ denotes the number of elements in the set $X$).
15. Let $\min _{1 \leqslant i \leqslant 20}\left\{\left|A_{i}\right|,\left|B_{i}\right|\right\}=t$.

Assume $\left|A_{1}\right|=t$,
$$
\begin{array}{l}
A_{1} \cap B_{i} \neq \varnothing(i=1,2, \cdots, k) ; \\
A_{1} \cap B_{j}=\varnothing(j=k+1, k+2, \cdots, 20) .
\end{array}
$$

Let $a_{i} \in A_{1} \cap B_{i}(i=1,2, \cdots, k)$.
By property (1), we know that $a_{1}, a_{2}, \cdots, a_{k}$ are distinct, $\left|A_{1}\right| \geqslant k$, i.e., $t \geqslant k$.

By property (2), we know that when $j=k+1, k+2, \cdots, 20$, $\left|A_{1}\right|+\left|B_{j}\right|=\left|A_{1} \cup B_{j}\right| \geqslant 18$, i.e.,
$$
\begin{array}{l}
\left|B_{j}\right| \geqslant 18-t . \\
\text { Then }|M|=\left|\bigcup_{i=1}^{20} B_{i}\right|=\sum_{j=1}^{20}\left|B_{j}\right| \\
\geqslant k t+(20-k)(18-t) \\
=360+2 k t-18 k-20 t \\
=180+2(k-10)(t-9) .
\end{array}
$$

If $t \leqslant 9$, then
$k \leqslant t \leqslant 9$,
$$
|M|=180+2(k-10)(t-9) \geqslant 180 \text {; }
$$

If $t \geqslant 10$, then $|M| \geqslant 20 t \geqslant 200$.
Therefore, $|M| \geqslant 180$ always holds.
Take $A_{i}=B_{i}=\{9(i-1)+j \mid j=1,2, \cdots, 9\}$, where $i=1,2, \cdots, 20$.

Then $M=\bigcup_{i=1}^{20} A_{i}=\bigcup_{i=1}^{20} B_{i}=\{1,2, \cdots, 180\}$ satisfies the requirement.
In this case, $|M|=180$.
In summary, the minimum number of elements in the set $M$ is 180.