11. The number of real solutions to the equation $\left(x^{2006}+1\right)\left(1+x^{2}+x^{4}+\cdots+x^{2004}\right)=2006 x^{2005}$ is
$$
\begin{array}{l}
\left(x^{2006}+1\right)\left(1+x^{2}+x^{4}+\cdots+x^{2004}\right)=\frac{\left(x^{2006}+1\right)\left(x^{2006}-1\right)}{x^{2}-1}=2006 x^{2005} \\
\Rightarrow x^{4012}-1=2016 x^{2005}\left(x^{2}-1\right)=\left(1+x^{2}+x^{4}+\cdots+x^{4010}\right)\left(x^{2}-1\right) .
\end{array}
$$

Upon inspection, $x=1$ is a root of the original equation, and $x=-1$ is not a root of the original equation.
When $x \neq \pm 1$, $1+x^{2}+x^{4}+\cdots+x^{4010}=2016 x^{2005} \geqslant 2016 \sqrt[2016]{x^{0+2+4+\cdots+4010}}$
$=2016|x|^{2005} \Rightarrow x \geqslant|x|$, no solution. Therefore, the number of real solutions to the original equation is 1.