China Mathematical Competition

We have $$\begin{array}{rl}& (x^{2006}+1)(1+x^2+x^4+\cdots +x^{2004})=2006x^{2005}\\ \iff & \left(x+\frac{1}{x^{2005}}\right)(1+x^2+x^4+\cdots +x^{2004})=2006\\ \iff & x+x^3+x^5+\cdots +x^{2005}+\frac{1}{x^{2005}}+\frac{1}{x^{2003}}+\frac{1}{x^{2001}}\\ & +\cdots +\frac{1}{x}=2006\\ \iff & 2006=x+\frac{1}{x}+x^3+\frac{1}{x^3}+\cdots +x^{2005}+\frac{1}{x^{2005}}\\ & \ge 2\times 1003=2006,\end{array}$$ where the equal holds if and only if $x=\frac{1}{x}$, $x^3=\frac{1}{x^3}$, ..., $x^{2005}=\frac{1}{x^{2005}}$. Then $x=\pm 1$. Since $x\le 0$ does not satisfy the original equation, $x=1$ is the only solution. So the number of real solutions is 1.