15. Given a quadratic equation in $x$
$$
\dot{x}^{2}-2 x-a^{2}-a=0 \quad (a>0) \text {. }
$$
(1) Prove: one root of this equation is greater than 2, and the other root is less than 2;
(2) If for $a=1,2, \cdots, 2004$, the two roots of the corresponding quadratic equations are $\alpha_{1}, \beta_{1}, \alpha_{2}, \beta_{2}$, $\cdots, \alpha_{2004} \times \beta_{2004}$, find
$$
\frac{1}{\alpha_{1}}+\frac{1}{\beta_{1}}+\frac{1}{\alpha_{2}}+\frac{1}{\beta_{2}}+\cdots+\frac{1}{\alpha_{2004}}+\frac{1}{\beta_{2004}}
$$

the value.
15. (1) Let the two roots of the equation be $x_{1}$ and $x_{2}$. By the relationship between roots and coefficients, we have
$$
\begin{array}{l}
x_{1}+x_{2}=2, x_{1} x_{2}=-a^{2}-a . \\
\text { Then }\left(2-x_{1}\right)\left(2-x_{2}\right)=4-2\left(x_{1}+x_{2}\right)+x_{1} x_{2} \\
=4-2 \times 2-a^{2}-a=-a^{2}-a<0 .
\end{array}
$$

Therefore, $2- x_{1}$ and $2-x_{2}$ have opposite signs, meaning that $x_{1}$ and $x_{2}$ are such that one is greater than 2 and the other is less than 2.
(2) When $a=1,2, \cdots, 2004$, the corresponding equations are
$$
\begin{array}{l}
x^{2}-2 x-2=0 ; \\
x^{2}-2 x-6=0 ; \\
\cdots \cdots . \\
x^{2}-2 x-2004 \times 2005=0 .
\end{array}
$$

By the relationship between roots and coefficients of a quadratic equation, we have
$$
\begin{array}{l}
\alpha_{1}+\beta_{1}=2, \alpha_{1} \beta_{1}=-2 ; \\
a_{2}+\beta_{2}=2, \alpha_{2} \beta_{2}=-6 ; \\
\text {..... } \\
a_{2000}+\beta_{2004}=2, \alpha_{200} \beta_{2000}=-2004 \times 2005 . \\
\text { Hence } \frac{1}{\alpha_{1}}+\frac{1}{\beta_{1}}+\frac{1}{a_{2}}+\frac{1}{\beta_{2}}+\cdots+\frac{1}{a_{2000}}+\frac{1}{\beta_{2} 000} \\
=\frac{\alpha_{1}+\beta_{1}}{\alpha_{1} \beta_{1}}+\frac{\alpha_{2}+\beta_{2}}{\alpha_{2} \beta_{2}}+\cdots+\frac{\alpha_{2004}+\beta_{2004}}{a_{2004} \beta_{2004}} \\
=-\frac{2}{2}-\frac{2}{6}-\cdots-\frac{2}{2004 \times 2005} \\
=-2\left(\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\cdots+\frac{1}{2004 \times 2006}\right) \\
=-2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdots+\frac{1}{2004}-\frac{1}{2000}\right) \\
=-2\left(1-\frac{1}{2005}\right)=-\frac{4008}{2005} . \\
\end{array}
$$