2003 AMC 10A Problem 18

Let $a = \dfrac{2003}{2004}.$ Multiplying the equation by $x$ gives $ax^2 + x + 1 = 0.$ If the roots are $r$ and $s,$ then by Vieta’s formulas $r + s = -\dfrac{1}{a}$ and $rs = \dfrac{1}{a}.$ The sum of reciprocals is $\dfrac{1}{r} + \dfrac{1}{s} = \dfrac{r + s}{rs} = \dfrac{-\frac{1}{a}}{\frac{1}{a}} = -1.$ Thus, the correct answer is B.