jmc

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 $$\begin{array}{rl}& x(x^2-2)(x^2-3)(x^2-4)+x(x^2-1)(x^2-3)(x^2-4)+x(x^2-1)(x^2-2)(x^2-4)+x(x^2-1)(x^2-2)(x^2-3)\\ & =2010(x^2-1)(x^2-2)(x^2-3)(x^2-4).\end{array}$$This expands to $$4x^7+\cdots =2010x^8+\cdots ,$$where only terms that have degree 7 or greater are shown. Then $$2010x^8-4x^7+\cdots =0,$$so by Vieta's formulas, the sum of the roots is $\frac{4}{2010}=\boxed{\frac{2}{1005}}.$