Four, on the ellipse $\frac{x^{2}}{5^{2}}+\frac{y^{2}}{4^{2}}=1$ there are 16 points, sequentially $P_{1}, P_{2}, \cdots, P_{16}, F$ is the left focus, and the angles between each adjacent pair of points and $F$ are equal $\left(\angle P_{1} F P_{2}=\angle P_{2} F P_{3}=\cdots=\angle P_{16} F P_{1}\right)$. Let the distance from $P_{i}$ to the left directrix be $d_{i}(i=1,2, \cdots, 16)$. Find $\sum_{i=1}^{16} \frac{1}{d_{i}}$.
$$
\begin{array}{l}
a=5, b \\
=4, c=3 . \text { Let } \angle X F P_{1} \\
=\alpha \cdot, \angle P_{1} F P_{2}= \\
\angle P_{2} F P_{3}=\cdots= \\
\angle P_{16} F P_{1}=\frac{\pi}{8} . \\
F M=d:-P_{i} F \cos \left[(i-1) \frac{\pi}{8}+\alpha\right] \\
\quad=\frac{a^{2}}{c}-c=\frac{16}{3} .
\end{array}
$$

By the definition of an ellipse, $P_{i} F=e d_{i}=\frac{c}{a} d_{i}=\frac{3}{5} d_{i}$, substituting into (1), we have
$$
\begin{array}{l}
d_{i}-\frac{3}{5} d_{i} \cos \left[\frac{(i-1) \pi}{8}+\alpha\right]=\frac{16}{3} . \\
\text { Thus } \frac{1}{d_{i}}=\frac{3}{16} \cdot \frac{1}{5}\left[5-3 \cos \left(\frac{(i-1) \pi}{8}+\alpha\right)\right] \\
\text { ( } i=1,2, \cdots, 16 \text { ). } \\
\therefore \sum_{i=1}^{16} \frac{1}{d}=\frac{3}{80}\left[\sum_{i=1}^{16} 5-3 \sum_{i=1}^{16} \cos \left(\frac{(i-1) \pi}{8}+\alpha\right)\right] \\
=3-\frac{9}{80} \sum_{i=1}^{16} \cos \left[\frac{(i-1) \pi}{8}+\alpha\right] \text {. } \\
\because \sum_{i=1}^{16} \cos \left[\frac{(i-1) \pi}{8}+\alpha\right] \\
=\sum_{i=1}^{i 6} \frac{2 \sin \frac{\pi}{16} \cos \left[(i-1)-\frac{\pi}{3}+\alpha\right]}{2 \sin \frac{\pi}{16}} \\
=\frac{1}{2 \sin \frac{\pi}{16}} \sum_{i=1}^{16}\left(\sin \left[\frac{\pi}{16}+(i-1) \frac{\pi}{8}+\alpha\right]\right. \\
\left.-\sin \left[(i-1) \frac{\pi}{8}+\alpha-\frac{\pi}{16}\right]\right) \\
=\frac{1}{2 \sin \frac{\pi}{16}}\left[\sin \left(\alpha+\frac{16 \pi}{8}-\frac{\pi}{16}\right)-\sin \left(x-\frac{\pi}{16}\right)\right] \\
=0 \text {, } \\
\therefore \sum_{i=1}^{16} \frac{1}{d_{i}}=3 \text {. } \\
\end{array}
$$