jmc

Substituting $\frac{1}{x}$, we have $$2f\left(\frac{1}{x}\right)+f\left(x\right)=\frac{5}{x}+4$$ This gives us two equations, which we can eliminate $f\left(\frac{1}{x}\right)$ from (the first equation multiplied by two, subtracting the second): $$\begin{array}{rl}3f(x)& =10x+4-\frac{5}{x}\\ 0& =x^2-\frac{3\times 2004-4}{10}x+\frac{5}{2}\end{array}$$ Clearly, the discriminant of the quadratic equation $\Delta >0$, so both roots are real. By Vieta's formulas, the sum of the roots is the coefficient of the $x$ term, so our answer is $\left[\frac{3\times 2004-4}{10}\right]=\boxed{601}$.