What is the sum of the reciprocals of the roots of the equation 
$\frac{2003}{2004}x+1+\frac{1}{x}=0$?
$\mathrm{(A) \ } -\frac{2004}{2003}\qquad \mathrm{(B) \ } -1\qquad \mathrm{(C) \ } \frac{2003}{2004}\qquad \mathrm{(D) \ } 1\qquad \mathrm{(E) \ } \frac{2004}{2003}$
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](https://artofproblemsolving.com/wiki/index.php/Vieta%27s_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 $\frac{a+b}{ab}=\frac{-\frac{2004}{2003}}{\frac{2004}{2003}}=-1 \Rightarrow\boxed{\mathrm{(B)}\ -1}$.