Four (18 points) As shown in Figure 3, $ABCD$ is a square with a side length of 1, $U$ and $V$ are points on $AB$ and $CD$ respectively, $AV$ intersects with $DU$ at point $P$, and $BV$ intersects with $CU$ at point $Q$. Find the maximum area of quadrilateral $PUQV$.
$$
\begin{array}{l}
\because A U ̈ / / D V, \\
\therefore S_{\triangle U P V} \\
=S_{\triangle U D V}-S_{\triangle P D V} \\
=S_{\triangle A D V}-S_{\triangle P D V} \\
=S_{\triangle A S P}. \\
\end{array}
$$

Similarly, $S_{\triangle U Q V}=S_{\triangle B Q C}$.
Therefore, $S_{\text {quadrilateral PUQV }}=S_{\triangle A P D}+S_{\triangle H S C}$.
Draw $P E \perp A D, Q F \perp B C$, with $E$ and $F$ as the feet of the perpendiculars, and let $P E = x, Q F = y$. Then
$$
S_{\text {quadrilateral }}=\frac{1}{2}(x+y).
$$

Let $A U = a, D V = b$. Then
$$
\frac{x}{a}+\frac{x}{b}=D E+A E=1.
$$

Thus, $x=\frac{a b}{a+b}$.
Similarly, $y=\frac{(1-a)(1-b)}{(1-a)+(1-b)}$
$$
=\frac{(1-a)(1-b)}{2-a-b}.
$$

Then $S_{\text {quadrilateral }}=\frac{1}{2}\left[\frac{a b}{a+b}+\frac{(1-a)(1-b)}{2-a-b}\right]$
$$
\begin{array}{l}
=\frac{(a+b)-\left(a^{2}+b^{2}\right)}{2(a+b)(2-a-b)} \\
=\frac{2(a+b)-a^{2}-b^{2}-\left(a^{2}+b^{2}\right)}{4(a+b)(2-a-b)} \\
\leqslant \frac{2(a+b)-a^{2}-b^{2}-2 a b}{4(a+b)(2-a-b)} \\
=\frac{(a+b)(2-a-b)}{4(a+b)(2-a-b)}=\frac{1}{4},
\end{array}
$$

The equality holds if and only if $a=b$.
Therefore, the maximum area of quadrilateral $P U Q V$ is $\frac{1}{4}$.