$8 \cdot 35$ The integer sequence $u_{0}, u_{1}, u_{2}, u_{3}, \cdots$ satisfies $u_{0}=1$, and for each positive integer $n$,
$$
u_{n+1} u_{n-1}=k u_{n},
$$

where $k$ is some fixed positive integer. If
$$
u_{2000}=2000,
$$

find all possible values of $k$.
Let $u_{1}=u$. By the given condition,
$$
u_{2}=\frac{k u_{1}}{u_{0}}=k u \text {. }
$$

If $u=0$, then $u_{1}=0, u_{2}=0$. If $u_{t-1}=0$, then $k u_{t}=u_{t-1} \cdot$ $u_{t+1}=0$, thus $u_{t}=0$. Therefore, for any positive integer $n, u_{n}=0$. This contradicts $u_{2000}=2000$. Hence, $u \neq 0$.
By the given condition, after calculation we get
$$
\begin{array}{ll}
u_{3}=\frac{k u_{2}}{u_{1}}=k^{\dot{2}}, & u_{4}=\frac{k u_{3}}{u_{2}}=\frac{k^{2}}{u}, \\
u_{5}=\frac{k u_{4}}{u_{3}}=\frac{k}{u}, & u_{6}=\frac{k u_{5}}{u_{4}}=1, \\
u_{7}=\frac{k u_{6}}{u_{5}}=u . &
\end{array}
$$

It can be seen that the given sequence is a periodic sequence with a period of 6. Therefore, by
$$
2000=6 \times 333+2
$$

we have
$$
u_{20000}=u_{2}=k u,
$$

Thus $k$ should satisfy
$$
\left\{\begin{array}{l}
k u=2000, \\
\frac{k}{u} \text { is an integer, }
\end{array}\right.
$$

where $k, u$ are both positive integers.
Substituting $k=u u_{5}$, we get
$$
\begin{array}{l}
u^{2} \cdot u_{5}=2000, \\
u^{2} \cdot u_{5}=2^{4} \cdot 5^{3} .
\end{array}
$$
The possible values of $u^{2}$ are $1,2^{2}, 2^{4}, 5^{2}, 2^{2} \cdot 5^{2}, 2^{4} 5^{2}$, thus the possible values of $u$ are 1, $2,2^{2}, 5,2 \cdot 5,2^{2} \cdot 5$; the possible values of $k$ are $2000,1000,500,400,200,100$.