imc

Multiplying both sides by $x$: $\frac{2003}{2004}{x}^2+1x+1=0$ Let the roots be $a$ and $b$. The problem is asking for $\frac{1}{a}+\frac{1}{b}=\frac{a+b}{ab}$ By Vieta's formulas: $a+b=(-1{)}^1\frac{1}{\frac{2003}{2004}}=-\frac{2004}{2003}$ $ab=(-1{)}^2\frac{1}{\frac{2003}{2004}}=\frac{2004}{2003}$ So the answer is $\frac{a+b}{ab}=\frac{-\frac{2004}{2003}}{\frac{2004}{2003}}=-1\Rightarrow \boxed{-1}$.