4. Let the complex roots of the equation $(3 x)^{2000}+(4 x+15)^{2000}=0$ be $\alpha_{i} 、 \overline{\alpha_{i}}(i=1,2, \cdots, 1000)$. Then $\sum_{i=1}^{1000} \frac{1}{\alpha_{i} a_{i}}$ equals ( ).
(A) $\frac{1000}{3}$
(B) $\frac{2000}{3}$
(C) $\frac{1000}{9}$
(D) $\frac{2000}{9}$
4.C.
The original equation is equivalent to $\left(\frac{4 x+15}{3 x}\right)^{2000}=-1$.
Let the 2000th roots of -1 be $\omega_{i} 、 \overline{\omega_{i}}(i=1,2$, $\cdots, 1000)$. Therefore,
$$
\sum_{i=1}^{1000}\left(\omega_{i}+\overline{\omega_{i}}\right)=0 \text {, and } \omega_{i}=\frac{4 \alpha_{i}+15}{3 \alpha_{i}} \text {. }
$$

Thus, $\frac{1}{\alpha_{i}}=\frac{\omega_{i}}{5}-\frac{4}{15}$,
$$
\begin{array}{l}
\frac{1}{\alpha_{i} \overline{a_{i}}}=\left(\frac{\omega_{i}}{5}-\frac{4}{15}\right)\left(\frac{\overline{\omega_{i}}}{5}-\frac{4}{15}\right) \\
=\frac{1}{25}+\frac{16}{25}-\frac{4}{15}\left(\frac{\omega_{i}}{5}+\frac{\overline{\omega_{i}}}{5}\right)=\frac{1}{9}-\frac{4}{75}\left(\omega_{i}+\overline{\omega_{i}}\right) . \\
\text { Hence } \sum_{i=1}^{1000} \frac{1}{\alpha_{i} \overline{a_{i}}}=\sum_{i=1}^{1000} \frac{1}{9}-\sum_{i=1}^{1000} \frac{4}{75}\left(\omega_{i}+\overline{\omega_{i}}\right) \\
=\frac{1000}{9} .
\end{array}
$$