We can add 4 to both sides to get
\[\frac{1}{x^2 - 1} + 1 + \frac{2}{x^2 - 2} + 1 + \frac{3}{x^2 - 3} + 1 + \frac{4}{x^2 - 4} + 1 = 2010x.\]This simplifies to
\[\frac{x^2}{x^2 - 1} + \frac{x^2}{x^2 - 2} + \frac{x^2}{x^2 - 3} + \frac{x^2}{x^2 - 4} = 2010x.\]We see that $x = 0$ is a solution (which does not affect our sum).  Otherwise, we can divide both sides by $x$:
\[\frac{x}{x^2 - 1} + \frac{x}{x^2 - 2} + \frac{x}{x^2 - 3} + \frac{x}{x^2 - 4} = 2010.\]Clearing the denominators, we get
\begin{align*}
&x(x^2 - 2)(x^2 - 3)(x^2 - 4) + x(x^2 - 1)(x^2 - 3)(x^2 - 4) + x(x^2 - 1)(x^2 - 2)(x^2 - 4) + x(x^2 - 1)(x^2 - 2)(x^2 - 3) \\
&\quad = 2010(x^2 - 1)(x^2 - 2)(x^2 - 3)(x^2 - 4).
\end{align*}This expands to
\[4x^7 + \dotsb = 2010x^8 + \dotsb,\]where only terms that have degree 7 or greater are shown.  Then
\[2010x^8 - 4x^7 + \dotsb = 0,\]so by Vieta's formulas, the sum of the roots is $\frac{4}{2010} = \boxed{\frac{2}{1005}}.$
Question:
Find the sum of all complex solutions of the equation
\[\frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} = 2010x - 4.\]
Answer:
Let's think step by step.

user
\frac{2}{1005}
Find the sum of all complex solutions of the equation
\[\frac{1}{x^2 - 1} + \frac{2}{x^2 - 2} + \frac{3}{x^2 - 3} + \frac{4}{x^2 - 4} = 2010x - 4.\]
test