jmc

By the given, $$2004(x^3-3x{y}^2)-2005(y^3-3x^{2}y)=0.$$Dividing both sides by $y^3$ and setting $t=\frac{x}{y}$ yields $$2004(t^3-3t)-2005(1-3t^2)=0.$$A quick check shows that this cubic has three real roots. Since the three roots are precisely $\frac{x_1}{y_1}$, $\frac{x_2}{y_2}$, and $\frac{x_3}{y_3}$, we must have $$2004(t^3-3t)-2005(1-3t^2)=2004\left(t-\frac{x_1}{y_1}\right)\left(t-\frac{x_2}{y_2}\right)\left(t-\frac{x_3}{y_3}\right).$$Therefore, $$\left(1-\frac{x_1}{y_1}\right)\left(1-\frac{x_2}{y_2}\right)\left(1-\frac{x_3}{y_3}\right)=\frac{2004(1^3-3(1))-2005(1-3(1{)}^2)}{2004}=\boxed{\frac{1}{1002}}.$$