3.2005 real numbers $x_{1}, x_{2}, \cdots, x_{2000}$ satisfy
$$
\begin{array}{l}
\left|x_{1}-x_{2}\right|+\left|x_{2}-x_{3}\right|+\cdots+ \\
\left|x_{2004}-x_{2000}\right|+\left|x_{2000}-x_{1}\right|=1 .
\end{array}
$$

Then $\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{2000}\right|$ has a minimum value of $\qquad$.
3. $\frac{1}{2}$.

For any positive integers $i, j$, we always have
$$
\left|x_{i}-x_{j}\right| \leqslant\left|x_{i}\right|+\left|x_{j}\right| \text { . }
$$
$$
\begin{array}{c}
\text { Then } 1=\left|x_{1}-x_{2}\right|+\left|x_{2}-x_{3}\right|+\cdots+ \\
\left|x_{2} \cos -x_{2} \cos \right|+\left|x_{2000}-x_{1}\right| \\
\leqslant\left|x_{1}\right|+\left|x_{2}\right|+\left|x_{2}\right|+\left|x_{3}\right|+\cdots+ \\
\left|x_{2004}\right|+\left|x_{2} \cos \right|+\left|x_{2} \cos \right|+\left|x_{1}\right| \\
=2\left(\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{2004}\right|+\left|x_{2} \cos \right|\right) .
\end{array}
$$

Therefore, $\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{2} \cos \right| \geqslant \frac{1}{2}$.
In fact, when $x_{1}=\frac{1}{2}, x_{2}=x_{3}=\cdots=x_{2000}=0$,
we have
$$
\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{2} \cos \right|=\frac{1}{2} .
$$

Therefore, the minimum value of $\left|x_{1}\right|+\left|x_{2}\right|+\cdots+\left|x_{2 \text { cos }}\right|$ is $\frac{1}{2}$.