Three, (25 points) 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 $\mid x_{1}+x_{2}+\cdots+$ $x_{2004} \mid$.
Three, from the known we can get
$$
\left\{\begin{array}{l}
x_{1}^{2}=x_{0}^{2}+2 x_{0}+1, \\
x_{2}^{2}=x_{1}^{2}+2 x_{1}+1, \\
\cdots \cdots . \\
x_{2004}^{2}=x_{2003}^{2}+2 x_{2003}+1 .
\end{array}\right.
$$

Thus, $x_{2004}^{2}=x_{0}^{2}+2\left(x_{0}+x_{1}+\cdots+x_{2003}\right)+2004$.
Given $x_{0}=0$, then
$$
\begin{array}{l}
2\left(x_{1}+x_{2}+\cdots+x_{2004}\right) \\
=x_{2004}^{2}+2 x_{2004}-2004=\left(x_{2004}+1\right)^{2}-2005,
\end{array}
$$

which means $\left.\left|x_{1}+x_{2}+\cdots+x_{2001}\right|=\frac{1}{2} \right\rvert\,\left(x_{2001}+1\right)^{2}-20051$.
Since $x_{1}+x_{2}+\cdots+x_{2004}$ is an integer, $x_{2004}+1$ must be an odd number. By comparing $143^{2}-20051$ and $145^{2}-20051$, we get
$$
\left|x_{1}+x_{2}+\cdots+x_{2004}\right| \geqslant \frac{1}{2}\left|45^{2}-2005\right|=10 \text {. }
$$

When $x_{0}=x_{2}=\cdots=x_{1900}=0, x_{1}=x_{3}=\cdots=x_{1999}=$ $-1, x_{11}=1, x_{1962}=2, x_{193}=3, \cdots, x_{2004}=44$, the equality holds.
Therefore, the minimum value of $\left|x_{1}+x_{2}+\cdots+x_{2004}\right|$ is 10.