20 Find the real solutions of the equation $\left(x^{2008}+1\right)\left(1+x^{2}+x^{4}+\cdots+x^{2006}\right)=2008 x^{2007}$.
(20 Divide both sides of the equation by $x^{2007}$, we get
$$
\begin{array}{c}
\left(x+\frac{1}{x^{2007}}\right)\left(1+x^{2}+x^{4}+\cdots+x^{2006}\right)=2008 \\
x+x^{3}+x^{5}+\cdots+x^{2007}+\frac{1}{x^{2007}}+\frac{1}{x^{2005}}+\cdots+\frac{1}{x}=2008, \\
2008=x+x^{3}+x^{5}+\cdots+x^{2007}+\frac{1}{x^{2007}}+\frac{1}{x^{2005}}+\cdots+\frac{1}{x} \\
\geqslant 2 \times 1004 \\
=2008
\end{array}
$$

The equality holds if and only if $x=\frac{1}{x}, x^{3}=\frac{1}{x^{3}}, \cdots, x^{2007}=\frac{1}{x^{2007}}$, which means $x=\pm 1$. However, when $x=-1$, it does not satisfy the original equation. Therefore, $x=1$ is the only real solution to the original equation.