9. (16 points) Given a positive integer $n(n \geqslant 3)$. It is known that $a_{1}<a_{2}<\cdots<a_{n}$, and $(k+1) a_{k} \in A(k=1,2$, $\left.\cdots, n-1, A=\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}\right)$. Find all ordered tuples $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ that satisfy the conditions.
9. First, $a_{1} \geqslant 0$. Otherwise, if $a_{1}<0$, then $2 a_{1}<a_{1}$. Therefore, it must be that
$$
a_{2}=2 a_{1}, a_{3}=3 a_{2}, \cdots, a_{n}=n a_{n-1} \text {. }
$$

Let $a_{1}=c$, then for $k \in\{2,3, \cdots, n\}$, we have
$$
a_{k}=k a_{k-1}=k(k-1) a_{k-2}=\cdots=k!a_{1}=k!c \text {. }
$$
(2) If $a_{1}=0$, then $a_{2}>0$. In this case, in the set $B$, $2 a_{1}=0,3 a_{2}>a_{2}$, and combining $B \subseteq A$, we know
$$
a_{3}=3 a_{2}, a_{4}=4 a_{3}, \cdots, a_{n}=n a_{n-1} \text {. }
$$

Let $a_{2}=d$, then for $k \in\{3,4, \cdots, n\}$, we have
$$
\begin{array}{l}
a_{k}=k a_{k-1}=k(k-1) a_{k-2}=\cdots \\
=k(k-1) \cdots \cdots 3 a_{2}=\frac{k!d}{2} .
\end{array}
$$

In summary, the ordered tuple $\left(a_{1}, a_{2}, \cdots, a_{n}\right)$ that satisfies the conditions is
$$
(c, 2!c, \cdots, n!c) \text { or }\left(0, \frac{2!d}{2}, \frac{3!d}{2}, \cdots, \frac{n!d}{2}\right),
$$

where $c$ and $d$ are any given positive numbers.