Four. (20 points) As shown in Figure 2, given that $A$ and $B$ are the left and right vertices of the ellipse $\Gamma$: $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$, and the line $l$ intersects the ellipse $\Gamma$ at points $M$ and $N$. Let the slopes of $A M$ and $B N$ be $k_{1}$ and $k_{2}$, respectively, and $k_{1}: k_{2}=1: 9$.
(1) Prove that the line $l$ passes through a fixed point;
(2) Let the areas of $\triangle A M N$ and $\triangle B M N$ be $S_{1}$ and $S_{2}$, respectively. Find the maximum value of $S_{1}-S_{2}$.
(1) As shown in Figure 4.
Introduce the scaling transformation \( T: (x, y) \rightarrow \left(x, \frac{5}{3} y\right) \).

Then the image of the ellipse \(\Gamma\) under the transformation \(T\) is a circle:
\[
\begin{array}{l}
x^{2} + y^{2} = 25. \\
\text{By } \frac{k_{1}}{k_{2}} = \frac{1}{9} \Rightarrow \frac{k_{A^{\prime} W^{\prime}}}{k_{B^{\prime} N^{\prime}}} = \frac{\frac{5}{3} k_{1}}{\frac{5}{3} k_{2}} = \frac{1}{9}.
\end{array}
\]

Let \(M^{\prime} N^{\prime}\) intersect the \(x\)-axis at point \(G^{\prime}\).
\[
\begin{array}{l}
\text{Then } \frac{1}{9} = \frac{k_{A^{\prime} X^{\prime}}}{k_{B^{\prime} N^{\prime}}} = \frac{\tan \angle M^{\prime} A^{\prime} B^{\prime}}{\tan \angle N^{\prime} B^{\prime} A^{\prime}} \\
= \frac{\left|M^{\prime} B^{\prime}\right|}{\left|M^{\prime} A^{\prime}\right|} \cdot \frac{\left|N^{\prime} B^{\prime}\right|}{\left|N^{\prime} A^{\prime}\right|} = \frac{\left|M^{\prime} B^{\prime}\right|}{\left|N^{\prime} A^{\prime}\right|} \cdot \frac{\left|N^{\prime} B^{\prime}\right|}{\left|M^{\prime} A^{\prime}\right|} \\
= \frac{\left|B^{\prime} G^{\prime}\right|}{\left|G^{\prime} N^{\prime}\right|} \cdot \frac{\left|G^{\prime} N^{\prime}\right|}{\left|A^{\prime} G^{\prime}\right|} = \frac{\left|B^{\prime} G^{\prime}\right|}{\left|A^{\prime} G^{\prime}\right|}.
\end{array}
\]

Therefore, point \(G^{\prime}(4,0)\), and line \(l\) passes through the fixed point \(G\).
(2) Suppose \(F^{\prime}(-4,0)\). Then
\[
\begin{array}{l}
S_{1} - S_{2} = \frac{3}{5} S_{\triangle A^{\prime} M^{\prime} N^{\prime}} - \frac{3}{5} S_{\triangle B^{\prime} M^{\prime} N^{\prime}} \\
= \frac{3}{5} \left(S_{\triangle A^{\prime} M^{\prime} N^{\prime}} - S_{\triangle B^{\prime} M^{\prime} N^{\prime}}\right) \\
= \frac{3}{5} S_{\triangle F^{\prime} M^{\prime} N^{\prime}} = \frac{6}{5} S_{\triangle O^{\prime} M^{\prime} N^{\prime}} \\
= \frac{6}{5} \times \frac{1}{2} \left|O^{\prime} M^{\prime}\right| \left|O^{\prime} N^{\prime}\right| \sin \angle M^{\prime} O^{\prime} N^{\prime} \\
\leqslant \frac{6}{5} \times \frac{1}{2} \times 5 \times 5 \sin 90^{\circ} = 15,
\end{array}
\]

with equality holding if and only if \(O^{\prime} M^{\prime} \perp O^{\prime} N^{\prime}\).
Therefore, the maximum value of \(S_{1} - S_{2}\) is 15.