What is the sum of the reciprocals of the roots of the equation \( \frac{2003}{2004} x+1+\frac{1}{x}=0 \) ? \( \begin{array}{ll}\text { (A) }-\frac{2004}{2003} & \text { (B) }-1 \\ \text { (C) } \frac{2003}{2004} & \text { (D) } 1\end{array} \)

To find the sum of the reciprocals of the roots of the equation \[ \frac{2003}{2004} x + 1 + \frac{1}{x} = 0, \] we start by multiplying through by \( x \) (assuming \( x \neq 0 \)) to eliminate the fraction: \[ \frac{2003}{2004} x^2 + x + 1 = 0. \] Next, we can rewrite this as: \[ \frac{2003}{2004} x^2 + x + 1 = 0. \] To find the sum of the reciprocals of the roots, we denote the roots of the quadratic equation as \( r_1 \) and \( r_2 \). The sum of the reciprocals of the roots can be expressed as: \[ \frac{1}{r_1} + \frac{1}{r_2} = \frac{r_1 + r_2}{r_1 r_2}. \] Using Vieta's formulas, we know that for a quadratic equation of the form \( ax^2 + bx + c = 0 \): - The sum of the roots \( r_1 + r_2 = -\frac{b}{a} \). - The product of the roots \( r_1 r_2 = \frac{c}{a} \). In our case, we have: - \( a = \frac{2003}{2004} \), - \( b = 1 \), - \( c = 1 \). Calculating the sum of the roots: \[ r_1 + r_2 = -\frac{b}{a} = -\frac{1}{\frac{2003}{2004}} = -\frac{2004}{2003}. \] Calculating the product of the roots: \[ r_1 r_2 = \frac{c}{a} = \frac{1}{\frac{2003}{2004}} = \frac{2004}{2003}. \] Now substituting these values into the expression for the sum of the reciprocals of the roots: \[ \frac{1}{r_1} + \frac{1}{r_2} = \frac{r_1 + r_2}{r_1 r_2} = \frac{-\frac{2004}{2003}}{\frac{2004}{2003}} = -1. \] Thus, the sum of the reciprocals of the roots is \[ \boxed{-1}. \]