Three, (50 points) The sequence of integers $a_{0}, a_{1}, a_{2}, a_{3}, \cdots$ satisfies $a_{0}=1$, and for each positive integer $n$, there is $a_{n+1} a_{n-1}$ $=k a_{n}$, where $k$ is a positive integer. If $a_{2000}=2000$, find all possible values of $k$.

保留源文本的换行和格式，直接输出翻译结果如下：

```
Three, (50 points) The sequence of integers $a_{0}, a_{1}, a_{2}, a_{3}, \cdots$ satisfies $a_{0}=1$, and for each positive integer $n$, there is $a_{n+1} a_{n-1}$ $=k a_{n}$, where $k$ is a positive integer. If $a_{2000}=2000$, find all possible values of $k$.
```
Let $a_{1}=a$, from the problem we know $a_{2}=k a_{1}=k a$.
If $a=0$, then $a_{2}=0$. Also, $a_{2} a_{4}=k a_{3}$, so $a_{3}=0$. By induction, we can get
$a_{n}=0\left(n \in \mathbf{N}_{+}\right)$. This contradicts $a_{2000}=2000$.
Therefore, $a \neq 0$. From the problem, we have
$$
a_{2}=k a, a_{3}=k^{2}, a_{4}=\frac{k^{2}}{a}, a_{5}=\frac{k}{a}, a_{6}=1, a_{7}=a .
$$

Thus, $a_{6 m+n}=a_{n}\left(n \in \mathbf{N}_{+}, n=0,1,2,3,4,5\right)$.
Since $2000=6 \times 333+2$, then $a_{2000}=a_{2}=2000=k a$.
Also, $k=a a_{5}$, so, $k a=a^{2} a_{5}=2000=2^{4} \times 5^{3}$.
Since $a_{5} \in \mathbf{Z}$, then $a^{2}$ is a positive divisor of 2000.
Hence $a^{2}=1,4,16,25,100,400$.
When $a=1$, from $k a=2000$ we know $k=2000$.
Similarly, $a=2, k=1000 ; a=4, k=500$;
$$
a=5, k=400 ; a=10, k=200 ; a=20, k=100 \text {. }
$$

Therefore, we get 6 possible values for $k$
$$
100,200,400,500,1000,2000 \text {. }
$$
(Ju Guoqi, Anhui Province Qianshan County No.2 Middle School, 246300)