5. Positive integers $a_{1}, a_{2}, \cdots, a_{18}$ satisfy the following two conditions:
(1) $a_{1}<a_{2}<a_{3} \cdots<a_{18} ;(2) a_{1}+a_{2}+\cdots+a_{18}=2001$, then the maximum value of $a_{9}$ is
5. 192
To make $a_{9}$ as large as possible, $a_{1}, a_{2}, \cdots, a_{8}$ should be as small as possible.
Therefore, take $a_{1}, a_{2}, \cdots, a_{8}$ to be $1,2, \cdots, 8$, and $1+2+\cdots+8=36$.
On the other hand, to make $a_{9}$ as large as possible, $a_{10}, a_{1}, \cdots, a_{18}$ should be as close to $a_{9}$ as possible.
Let $a_{9}=n$, then $\frac{(n+n+9) \times 10}{2}=2001-36$. Thus, $n=192$. Therefore, the maximum value of $a_{9}$ is 192.