We have
$$\begin{align*} & (x^{2006} + 1)(1 + x^2 + x^4 + \cdots + x^{2004}) = 2006x^{2005} \\ \Leftrightarrow & \left(x + \frac{1}{x^{2005}}\right)(1 + x^2 + x^4 + \cdots + x^{2004}) = 2006 \\ \Leftrightarrow & x + x^3 + x^5 + \cdots + x^{2005} + \frac{1}{x^{2005}} + \frac{1}{x^{2003}} + \frac{1}{x^{2001}} \\ & \qquad + \cdots + \frac{1}{x} = 2006 \\ \Leftrightarrow & 2006 = x + \frac{1}{x} + x^3 + \frac{1}{x^3} + \cdots + x^{2005} + \frac{1}{x^{2005}} \\ & \ge 2 \times 1003 = 2006, \end{align*}$$
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.