15. Let $M$ be a set composed of a finite number of positive integers, and $M=A_{1} \cup A_{2} \cup \cdots \cup A_{20}=B_{1} \cup B_{2} \cup \cdots \cup B_{20}$, where $A_{i} \neq \varnothing, B_{i} \neq \varnothing, i=1,2, \cdots, 20$, and for any $1 \leq i<j \leq 20$, we have $A_{i} \cap A_{j}=\varnothing, B_{i} \cap B_{j}=\varnothing$.

It is known that for any $1 \leq i \leq 20,1 \leq j \leq 20$, if $A_{i} \cap B_{j}=\varnothing$, then $\left|A_{i} \cup B_{j}\right| \geq 18$, find the minimum number of elements in the set $M$ (here, $|X|$ denotes the number of elements in the set $X$).
Let $\min \left\{\left|A_{1}\right|,\left|A_{2}\right|, \cdots,\left|A_{20}\right|,\left|B_{1}\right|,\left|B_{2}\right|, \cdots,\left|B_{20}\right|\right\}=t$.
Without loss of generality, assume $\left|A_{1}\right|=t, A_{1} \cap B_{i} \neq \varnothing, i=1,2, \cdots, k ; A_{1} \cap B_{j}=\varnothing, j=k+1, k+2, \cdots, 20$.
Let $a_{i} \in A_{1} \cap B_{i}, i=1,2, \cdots, k$.
Since for any $1 \leq i<j \leq 20$, we have $B_{i} \cap B_{j}=\varnothing$, it follows that $a_{1}, a_{2}, \cdots, a_{k}$ are distinct, $\left|A_{1}\right| \geq k$, i.e., $t \geq k$.
(5 points)

Since for any $1 \leq i \leq 20,1 \leq j \leq 20$, if $A_{i} \cap B_{j}=\varnothing$, then $\left|A_{i} \cup B_{j}\right| \geq 18$, so when $j=k+1, k+2, \cdots, 20$, $\left|A_{1}\right|+\left|B_{j}\right|=\left|A_{1} \cup B_{j}\right| \geq 18$; i.e., when $j=k+1, k+2, \cdots, 20$, $\left|B_{j}\right| \geq 18-t$.
Therefore,
$$
\begin{aligned}
|M| & =\left|B_{1} \cup B_{2} \cup \cdots \cup B_{20}\right|=\left|B_{1}\right|+\left|B_{2}\right|+\cdots+\left|B_{k}\right|+\left|B_{k+1}\right|+\cdots+\left|B_{20}\right| \\
& \geq k t+(20-k)(18-t)=360+2 k t-18 k-20 t=180+2(k-10)(t-9)
\end{aligned}
$$

If $t \leq 9$, then $k \leq t \leq 9,|M|=180+2(k-10)(t-9) \geq 180$; if $t \geq 10$, then $|M| \geq 20 t \geq 200$.
Therefore, we always have $|M| \geq 180$.
On the other hand, take $A_{i}=B_{i}=\{9(i-1)+1,9(i-1)+2, \cdots, 9(i-1)+9\}, i=1,2, \cdots, 20$, then
$$
M=A_{1} \cup A_{2} \cup \cdots \cup A_{20}=B_{1} \cup B_{2} \cup \cdots \cup B_{20}=\{1,2, \cdots, 180\}
$$

which meets the requirements, and in this case, $|M|=180$.
In conclusion, the minimum number of elements in the set $M$ is 180.
(20 points)