Four. (20 points) As shown in Figure 3, let $F$ be one of the foci of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$, and $A$ be the vertex of the ellipse that is farthest from $F$. Take 2008 distinct points $P_{i}(i=1,2, \cdots, 2008)$ on the minor axis $B C$ of the ellipse. Let $F P_{i}$ intersect $A B$ or $A C$ at point $M_{i}$, and $A P_{i}$ intersect $F B$ or $F C$ at point $N_{i}$. How many pieces do the lines $M_{i} N_{i}(i=1,2, \cdots, 2008)$ divide the ellipse into?
For any point $P_{i}(i=1,2, \cdots, 2008)$, by symmetry, we can assume $P_{i}$ is on the positive half of the $y$-axis, as shown in Figure 7.

In $\triangle A F C$, since $A N_{i}$, $C O$, and $F M_{i}$ are concurrent, by Ceva's Theorem, we have
$$
\frac{A M_{i}}{M_{i} C} \cdot \frac{C N_{i}}{N_{i} F} \cdot \frac{F O}{O A}=1 .
$$

Let the line $M_{i} N_{i}$ intersect the $x$-axis at point $Q_{i}(i=1,2, \cdots, 2008)$. $\triangle A C F$ is intersected by the line $Q_{i} N_{i} M_{i}$. By Menelaus' Theorem, we have
$$
\frac{A M_{i}}{M_{i} C} \cdot \frac{C N_{i}}{N_{i} F} \cdot \frac{F Q_{i}}{Q_{i} A}=1 .
$$

Dividing the two equations, we get $\frac{F O}{O A}=\frac{F Q_{i}}{Q_{i} A}=\frac{F Q_{i}}{Q_{i} F+F A}$.
Thus, $F Q_{i}=\frac{F A \cdot O F}{O A-O F}=\frac{a c+c^{2}}{a-c}=12>5=a$.
Therefore, $M_{i} N_{i}(i=1,2, \cdots, 2008)$ passes through a fixed point on the extension of the major axis of the ellipse.

Hence, the lines $M_{i} N_{i}(i=1,2, \cdots, 2008)$ divide the ellipse into $2008+1=2009$ parts.