11. The number of real solutions to the equation $\left(x^{2006}+1\right)\left(1+x^{2}+x^{4}+\cdots+\right.$ $\left.x^{2004}\right)=2006 x^{2005}$ is $\qquad$
11.1.
$$
\begin{array}{l}
\left(x^{2006}+1\right)\left(1+x^{2}+x^{4}+\cdots+x^{2004}\right)=2006 x^{2005} \\
\Leftrightarrow\left(x+\frac{1}{x^{2005}}\right)\left(1+x^{2}+x^{4}+\cdots+x^{2004}\right)=2006 \\
\Leftrightarrow x+x^{3}+x^{5}+\cdots+x^{2005}+\frac{1}{x^{2008}}+\frac{1}{x^{2003}} \\
\quad+\cdots+\frac{1}{x}=2006 \\
\Leftrightarrow x+\frac{1}{x}+x^{3}+\frac{1}{x^{3}}+\cdots+x^{2005}+\frac{1}{x^{2005}} \\
\geqslant 2 \times 1003=2006 .
\end{array}
$$

For the equality to hold, it must be that $x=\frac{1}{x}, x^{3}=\frac{1}{x^{3}}, \cdots, x^{2005}=\frac{1}{x^{2005}}$, i.e., $x= \pm 1$.
However, when $x \leqslant 0$, the original equation is not satisfied.
Therefore, $x=1$ is the only solution to the original equation.
Thus, the number of real solutions to the original equation is 1.