11. From the given, we have
$$
x_{n}^{2}=x_{n-1}^{2}+2 x_{n}+1(n=2,3, \cdots, 2014) \text {. }
$$

Adding the above 2013 equations, we get
$$
\begin{array}{l}
x_{2014}^{2}=x_{1}^{2}+2\left(x_{1}+x_{2}+\cdots+x_{2013}\right)+2013 \\
\Rightarrow 2\left(x_{1}+x_{2}+\cdots+x_{2014}\right) \\
\quad=x_{2014}^{2}+2 x_{2014}-2013-x_{1}^{2} \\
\quad=\left(x_{2014}+1\right)^{2}-2014-99^{2} .
\end{array}
$$

Since $x_{1}$ is odd, from the given conditions, we know that $x_{2}$ is even, $x_{3}$ is odd, $\cdots, x_{2014}$ is even.
Thus, $\left(x_{2014}+1\right)^{2} \geqslant 1$.
Therefore, $x_{1}+x_{2}+\cdots+x_{2014}$
$$
\geqslant \frac{1}{2}\left(1-2014-99^{2}\right)=-5907 \text {. }
$$

When $x_{1}=-99, x_{2}=-98, \cdots, x_{99}=-1, x_{100}$ $=0, x_{101}=-1, x_{102}=0, x_{103}=-1, \cdots, x_{2013}=$ $-1, x_{2014}=0$, the equality holds.
Hence, the minimum value of $x_{1}+x_{2}+\cdots+x_{2014}$ is -5907.
11. Let real numbers $x_{1}, x_{2}, \cdots, x_{2011}$ satisfy
$$
\left|x_{1}\right|=99,\left|x_{n}\right|=\left|x_{n-1}+1\right| \text {, }
$$

where, $n=2,3, \cdots, 2014$. Find the minimum value of $x_{1}+x_{2}+\cdots+x_{2014}$.