3. Given $x_{1}=x_{2011}=1$, $\left|x_{n+1}\right|=\left|x_{n}+1\right|(n=1,2, \cdots, 2010)$.
Then $x_{1}+x_{2}+\cdots+x_{2010}=$ $\qquad$
3. -1005 .

From the given, it is easy to obtain
$$
x_{n+1}^{2}=x_{n}^{2}+2 x_{n}+1(n=1,2, \cdots, 2010) \text {. }
$$

By summing up and organizing these 2010 equations, we get
$$
\begin{array}{l}
x_{2011}^{2}=2\left(x_{1}+x_{2}+\cdots+x_{2010}\right)+2011 . \\
\text { Therefore, } x_{1}+x_{2}+\cdots+x_{2010}=-1005 .
\end{array}
$$