8 Nine consecutive positive integers are arranged in ascending order to form a sequence $a_{1}, a_{2}, \cdots, a_{9}$. If the value of $a_{1}+a_{3}+a_{5}+a_{7}+a_{9}$ is a perfect square, and the value of $a_{2}+a_{4}+a_{6}+a_{8}$ is a perfect cube, then the minimum value of the sum of these nine positive integers is $\qquad$ .
818000 Hint: Let these nine numbers be $a-4, a-3, a-2, a-1, a, a+$ $1, a+2, a+3, a+4$, then we have, $5 a=m^{2}, 4 a=n^{3}, S=9 a$, so $a=$ $\frac{m^{2}}{5}=\frac{n^{2}}{4}$, which gives
$$
4 m^{2}=5 n^{3} .
$$

Let $n=2 n_{1}, m=5 m_{1}$, then $100 m_{1}^{2}=40 n_{1}^{3}$, so $5 m_{1}^{2}=2 n_{1}^{3}$. Taking $m_{1}=2 m_{2}, n_{1}=5 n_{2}$, it becomes $2 m_{2}^{2}=5^{2} n_{2}^{3}$. Taking $m_{2}=10, n_{2}=2$, the left side can be satisfied, at this time $n=20, m=100, a=2000, S=9 a=18000$.