11. 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 $k$.
11. Let $u_{1}=u$. By the given condition, $u_{2}=\frac{k u_{1}}{u_{0}}=k u$

If $u=0$, then $u_{1}=0, u_{2}=0$. If $u_{t-1}=0$, then $k u_{t}=u_{t-1} \cdot u_{t+1}=0$, hence $u_{t}=0$. Therefore, for any positive integer $n, u_{n}=0$. This contradicts $u_{2000}=2000$. Thus, $u \neq 0$.

By the given condition, after calculation, we get $u_{3}=\frac{k u_{2}}{u_{1}}=k^{2}, u_{4}=\frac{k u_{3}}{u_{2}}=\frac{k^{2}}{u}, u_{5}=\frac{k u_{4}}{u_{3}}=\frac{k}{u}, u_{6}=\frac{k u_{5}}{u_{4}}=1, u_{7}=\frac{k u_{6}}{u_{5}}=u$. Therefore, the given sequence is a periodic sequence with a period of 6. Since $2000=6 \times 333+2$, we have $u_{2000}=u_{2}=k u$. Therefore, $k$ should satisfy $\left\{\begin{array}{l}k u=2000, \\ \frac{k}{u} \text { is an integer, }\end{array}\right.$ where $k, u$ are positive integers. Thus, $u^{2} \cdot u_{5}=u\left(u \cdot u_{5}\right)=u k=2000=2^{4} \cdot 5^{3}$.
Hence, the possible values of $u^{2}$ are $1,2^{2}, 2^{4}, 5^{2}, 2^{2} \cdot 5^{2}, 2^{4} \cdot 5^{2}$, and the possible values of $u$ are $1,2,2^{2}, 5,2 \times 5,2^{2} \times 5$; the possible values of $k$ are $2000,1000,500,400,200,100$.