jmc

Note that $x=0$ is not a solution. Also, if $x<0,$ then the left-hand side is positive and the right-hand side is negative, so $x$ cannot be a solution. Thus, any real roots must be positive. Assume $x>0.$

Dividing both sides by $x^{2005},$ we get $$\frac{(x^{2006}+1)(x^{2004}+x^{2002}+x^{2000}+\cdots +x^2+1)}{x^{2005}}=2006.$$Then $$\frac{x^{2006}+1}{x^{1003}}\cdot \frac{x^{2004}+x^{2002}+x^{2000}+\cdots +x^2+1}{x^{1002}}=2006,$$or $$\left(x^{1003}+\frac{1}{x^{1003}}\right)\left(x^{1002}+x^{1000}+x^{998}+\cdots +\frac{1}{x^{998}}+\frac{1}{x^{1000}}+\frac{1}{x^{1002}}\right)=2006.$$By AM-GM, $$\begin{array}{rl}{x}^{1003}+\frac{1}{x^{1003}}& \ge 2,\\ x^{1002}+x^{1000}+x^{998}+\cdots +\frac{1}{x^{998}}+\frac{1}{x^{1000}}+\frac{1}{x^{1002}}& \ge \sqrt[1003]{x^{1002}\cdot x^{1000}\cdot x^{998}\cdots \frac{1}{x^{998}}\cdot \frac{1}{x^{1000}}\cdot \frac{1}{x^{1002}}}=1003,\end{array}$$so $$\left(x^{1003}+\frac{1}{x^{1003}}\right)\left(x^{1002}+x^{1000}+x^{998}+\cdots +\frac{1}{x^{998}}+\frac{1}{x^{1000}}+\frac{1}{x^{1002}}\right)\ge 2006.$$Since we have the equality case, the only possible value of $x$ is 1, so there is $\boxed{1}$ real root.