16. Nine consecutive positive integers are arranged in an increasing sequence $a_{1}, a_{2}, \cdots, a_{9}$. If $a_{1}+a_{3}+a_{5}+a_{7}+a_{9}$ is a perfect square, and $a_{2}+a_{4}+a_{6}+a_{8}$ is a perfect cube, then the minimum value of $a_{1}+a_{2}+\cdots+a_{9}$ is
16. 18000.

Let these nine numbers be $a-4, a-3, \cdots, a+4$. Then $5 a=m^{2}, 4 a=n^{3}, S=9 a$.
Thus, $a=\frac{m^{2}}{5}=\frac{n^{3}}{4} \Rightarrow 4 m^{2}=5 n^{3}$. Clearly, $m$ and $n$ are both multiples of 10.
Let $m=10 m_{1}, n=10 n_{1}$, then $4 m_{1}^{2}=50 n_{1}^{3}$.
Take $m_{1}=10, n_{1}=2$. At this point,
$$
\begin{array}{l}
n=20, m=100 \\
\Rightarrow a=2000, S=9 a=18000 .
\end{array}
$$