16. Through the right focus $F$ of the ellipse $\frac{x^{2}}{3}+\frac{y^{2}}{2}=1$, draw two perpendicular chords $A B$ and $C D$. Let the midpoints of $A B$ and $C D$ be $M$ and $N$ respectively.
(1) Prove that the line $M N$ must pass through a fixed point, and find this fixed point;
(2) If the slopes of chords $A B$ and $C D$ both exist, find the maximum value of the area $S$ of $\triangle F M N$.
16. (1) From the problem, we know $F(1,0)$.
(i) When the slopes of the chords $A B$ and $C D$ both exist, let the slope of $A B$ be $k$. Then the slope of $C D$ is $-\frac{1}{k}$.
Let $l_{A B}: y=k(x-1)$.
Substitute into the ellipse equation $\frac{x^{2}}{3}+\frac{y^{2}}{2}=1$, we get
$$
\left(3 k^{2}+2\right) x^{2}-6 k^{2} x+\left(3 k^{2}-6\right)=0 \text {. }
$$

Thus, $x_{M}=\frac{x_{A}+x_{B}}{2}=\frac{3 k^{2}}{3 k^{2}+2}$,
$$
y_{M}=k\left(x_{M}-1\right)=\frac{-2 k}{3 k^{2}+2} \text {. }
$$

Therefore, $M\left(\frac{3 k^{2}}{3 k^{2}+2}, \frac{-2 k}{3 k^{2}+2}\right)$.
Since $C D \perp A B$, by substituting $k$ with $-\frac{1}{k}$ in the coordinates of point $M$, we get point $N\left(\frac{3}{2 k^{2}+3}, \frac{2 k}{2 k^{2}+3}\right)$.
1) When $k \neq \pm 1$,
$$
k_{M N}=\frac{\frac{2 k}{2 k^{2}+3}+\frac{2 k}{3 k^{2}+2}}{\frac{3}{2 k^{2}+3}-\frac{3 k^{2}}{3 k^{2}+2}}
=\frac{10 k\left(k^{2}+1\right)}{6-6 k^{4}}=\frac{-5 k}{3 k^{2}-3}.
$$
At this time, $l_{M N}: y-\frac{2 k}{2 k^{2}+3}=\frac{-5 k}{3 k^{2}-3}\left(x-\frac{3}{2 k^{2}+3}\right)$.
Then the line $l_{M N}$ passes through the fixed point $\left(\frac{3}{5}, 0\right)$.
2) When $k= \pm 1$, it is easy to get $l_{M N}: x=\frac{3}{5}$, which also passes through the point $\left(\frac{3}{5}, 0\right)$.
(ii) When the slope of chord $A B$ or chord $C D$ does not exist, it is easy to know that the line $M N$ is the $x$-axis, which also passes through the point $\left(\frac{3}{5}, 0\right)$.
In summary, the line $M N$ must pass through the fixed point $E\left(\frac{3}{5}, 0\right)$.
(2) From (1), we know
$$
\begin{aligned}
S & =\frac{1}{2}|E F|\left|y_{M}-y_{N}\right| \\
& =\frac{1}{5}\left|\frac{-2 k}{3 k^{2}+2}-\frac{2 k}{2 k^{2}+3}\right| \\
& =\frac{\left|2 k\left(k^{2}+1\right)\right|}{\left(3 k^{2}+2\right)\left(2 k^{2}+3\right)} .
\end{aligned}
$$

Assume $k>0$. Then by differentiation, we get
$$
\begin{aligned}
S^{\prime} & =\frac{-12 k^{6}-10 k^{4}+10 k^{2}+12}{\left(3 k^{2}+2\right)^{2}\left(2 k^{2}+3\right)^{2}} \\
& =\frac{\left(-12 k^{4}-22 k^{2}-12\right)\left(k^{2}-1\right)}{\left(3 k^{2}+2\right)^{2}\left(2 k^{2}+3\right)^{2}} .
\end{aligned}
$$

From $S^{\prime}=0$, we know $k=1$.
Also, when $k \in(0,1)$, $S^{\prime}>0$;
when $k \in(1,+\infty)$, $S^{\prime}<0$.
Therefore, when $k=1$, $S$ has a maximum value of $\frac{4}{25}$.
Thus, the maximum area of $\triangle F M N$ is $\frac{4}{25}$.