9. From the known, we have
$$
\left\{\begin{array}{l}
x_{1}^{2}=x_{3}^{2}+2 x_{0}+1, \\
x_{2}^{2}=x_{1}^{2}+2 x_{1}+1, \\
\cdots \cdots \\
x_{2004}^{2}=x_{2003}^{2}+2 x_{2003}+1 .
\end{array}\right.
$$

Thus, $x_{2004}^{2}=x^{2}+2\left(x_{0}+x_{1}+\cdots+x_{2003}\right)+2004$. Given $x_{0}=0$, then $2\left(x_{1}+x_{2}+\cdots+x_{2004}\right)=x_{2004}^{2}+2 x_{2004}$ $-2004=\left(x_{2004}+1\right)^{2}-2005$, hence $\left|x_{1}+x_{2}+\cdots+x_{2004}\right|=\frac{1}{2}\left|\left(x_{2004}+1\right)^{2}-2005\right|$. Since $x_{1}+x_{2}+\cdots+x_{2004}$ is an integer, $x_{2004}+1$ must be an odd number. By comparing $\left|43^{2}-2005\right|$ and $\left|45^{2}-2005\right|$, we find that $\left|x_{1}+x_{2}+\cdots+x_{2004}\right| \geqslant \frac{1}{2}$ $\left|45^{2}-2005\right|=10$. When $x_{0}=x_{2}=\cdots=x_{1960}=0, x_{1}=x_{3}=\cdots=x_{1959}=-1, x_{1961}=1, x_{1962}=2, x_{1963}=3$, $\cdots, x_{2004}=44$, the equality holds. Therefore, the minimum value of $\left|x_{1}+x_{2}+\cdots+x_{2004}\right|$ is 10.
9. The integers $x_{0}, x_{1}, \cdots, x_{2000}$ satisfy the conditions, $x_{0}=0,\left|x_{1}\right|=\left|x_{0}+1\right|,\left|x_{2}\right|=\left|x_{1}+1\right|, \cdots$, $\left|x_{2004}\right|=\left|x_{2003}+1\right|$, find the minimum value of $\left|x_{1}+x_{2}+\cdots+x_{2004}\right|$.