4. Given a convex 20-gon $P$. Use 17 non-intersecting diagonals of $P$ to divide $P$ into 18 triangles, and the resulting figure is called a triangulation of $P$. For any triangulation $T$ of $P$, the 20 sides of $P$ and the added 17 diagonals are collectively referred to as the edges of $T$. A set of 10 edges of $T$ that have no common endpoints is called a perfect matching of $T$. When $T$ ranges over all triangulations of $P$, find the maximum number of perfect matchings of $T$.
Consider the general problem by replacing the 20-gon with a $2n$-gon. For a diagonal of a convex $2n$-gon $P$, if there are an odd number of $P$'s vertices on each side, it is called an odd chord; otherwise, it is called an even chord. First, note the following basic fact:
Lemma 1: For any triangulation graph $T$ of $P$, the perfect matching of $T$ does not contain odd chords.
If there is an odd chord $e_1$ in the perfect matching, because a perfect matching of $T$ gives a partition of the vertex set of $P$, and there are an odd number of vertices on each side of $e_1$, there must be another edge $e_2$ of $T$ in the perfect matching, with endpoints on different sides of $e_1$. Since $P$ is a convex polygon, $e_1$ and $e_2$ intersect inside $P$, which contradicts the fact that $T$ is a triangulation graph.
Let $f(T)$ be the number of perfect matchings of $T$. Let $F_1 = 1, F_2 = 2$, and for $k \geq 2, F_{k+2} = F_{k+1} + F_k$ be the Fibonacci sequence. We will prove by induction on $n$:
Lemma 2: If $T$ is any triangulation graph of a convex $2n$-gon, then $f(T) \leq F_n$. Let $P = A_1 A_2 \cdots A_{2n}$ be a convex $2n$-gon. Selecting $n$ edges from the $2n$ edges of $P$ to form a perfect matching, there are exactly two ways: $A_1 A_2, A_3 A_4, \cdots, A_{2n-1} A_{2n}$ or $A_2 A_3, A_4 A_5, \cdots, A_{2n} A_1$. When $n=2$, for the triangulation graph $T$ of the convex quadrilateral $P$, there are no even chords, so the perfect matching can only use the edges of $P$, i.e., $f(T) = 2 = F_2$.
When $n=3$, for the triangulation graph $T$ of the convex hexagon $P$, there is at most one even chord. If $T$ has no even chords, as above, $f(T) = 2$. If $T$ contains an even chord, assume it is $A_1 A_4$, then the perfect matching using $A_1 A_4$ is unique, and the other two edges can only be $A_2 A_3, A_5 A_6$, so $f(T) = 3$. In summary, $f(T) \leq 3 = F_3$. The conclusion holds for $n=2,3$;
Assume $n \geq 4$, and the conclusion holds for all values less than $n$. Consider a triangulation graph $T$ of the convex $2n$-gon $A_1 A_2 \cdots A_{2n}$. If $T$ has no even chords, as above, $f(T) = 2$.
For an even chord $e$, let the smaller number of vertices of $P$ on either side of $e$ be $w(e)$. If $T$ contains an even chord, take an even chord $e$ such that $w(e)$ is minimized. Let $w(e) = 2k$, and assume $e$ is $A_{2n} A_{2k+1}$, then no $A_i (i=1,2, \cdots, 2k)$ can have an even chord.
In fact, if $A_i A_j$ is an even chord, if $j \in \{2k+2, 2k+3, \cdots, 2n-1\}$, then $A_i A_j$ intersects $e$ inside $P$, a contradiction; if $j \in \{1,2, \cdots, 2k+1, 2n\}$, then $w(A_i A_j) < 2k$, contradicting the minimality of $w(e)$.
By Lemma 1, the perfect matching does not contain odd chords, so $A_1, A_2, \cdots, A_{2k}$ can only be paired with their adjacent vertices, and in particular, $A_1$ can only be paired with $A_2$ or $A_{2n}$. We consider two cases.
Case 1: Use the edge $A_1 A_2$, then the edges $A_3 A_4, \cdots, A_{2k-1} A_{2k}$ must be used. Note that $A_{2n} A_{2k+1}$ has $2k$ and $2n-2k-2$ vertices on either side, and $2n-2k-2 \geq w(A_{2n} A_{2k+1}) = 2k$, and since $n \geq 4$, $2n-2k \geq 6$. In the convex $2n-2k$-gon $P_1 = A_{2k+1} A_{2k+2} \cdots A_{2n}$, the edges of $T$ form a triangulation graph $T_1$ of $P_1$. Selecting $n-k$ edges $e_1, e_2, \cdots, e_{n-k}$ from $T$, together with $A_1 A_2, A_3 A_4, \cdots, A_{2k-1} A_{2k}$, forms a perfect matching of $T$ if and only if $e_1, e_2, \cdots, e_{n-k}$ is a perfect matching of $T_1$. In this case, the number of perfect matchings of $T$ is equal to $f(T_1)$.
Case 2: Use the edge $A_1 A_{2n}$, then the edges $A_2 A_3, \cdots, A_{2k} A_{2k+1}$ must be used. In the convex $2n-2k-2$-gon $P_2 = A_{2k+2} A_{2k+3} \cdots A_{2n-1}$, construct a triangulation graph $T_2$ as follows: for $2k+2 \leq i < j \leq 2n-1$, if the segment $A_i A_j$ is an edge of $T$, then it is also an edge of $T_2$. Since these edges do not intersect inside, we can add some diagonals of $P_2$ to form a triangulation graph $T_2$ of $P_2$ that includes all the edges between the vertices $A_{2k+2}, A_{2k+3}, \cdots, A_{2n-1}$ of $T$. Thus, each perfect matching of $T$ that includes the edges $A_1 A_2, A_3 A_4, \cdots, A_{2k-1} A_{2k}$ must have the remaining edges as a perfect matching of $T_2$. In this case, the number of perfect matchings of $T$ is at most $f(T_2)$.
By the induction hypothesis, $f(T_1) \leq F_{n-k}$ and $f(T_2) \leq F_{n-k-1}$. Combining the two cases and noting that $k \geq 1$, we have $f(T) \leq f(T_1) + f(T_2) \leq F_{n-k} + F_{n-k-1} = F_{n-k+1} \leq F_n$.
We now show that the equality can be achieved.
Consider the triangulation graph $\Delta_n$ of the convex $2n$-gon $A_1 A_2 \cdots A_{2n}$:
Add diagonals $A_2 A_{2n}, A_{2n} A_3, A_3 A_{2n-1}, A_{2n-1} A_4, A_4 A_{2n-2}, \cdots, A_{n+3} A_n, A_n A_{n+2}$.
Repeating the previous argument, $f(\Delta_2) = 2$ and $f(\Delta_3) = 3$.

For $\Delta_n, n \geq 4$, consider the even chord $A_3 A_{2n}$.
Case 1: Use $A_1 A_2$. Since the triangulation graph of the convex $2n-2$-gon $A_3 A_4 \cdots A_{2n}$ is exactly $\Delta_{n-1}$, there are $f(\Delta_{n-1})$ perfect matchings in this case; Case 2: Use $A_1 A_{2n}$. Since the triangulation graph of the convex $2n-4$-gon $A_4 A_5 \cdots A_{2n-1}$ is exactly $\Delta_{n-2}$, no additional diagonals need to be added, and the number of perfect matchings in this case is exactly $f(\Delta_{n-2})$.
Thus, for $n \geq 4$, we have $f(\Delta_n) = f(\Delta_{n-1}) + f(\Delta_{n-2})$. By mathematical induction, $f(\Delta_n) = F_n$. The conclusion is proved. Therefore, for the convex 20-gon $P$, the maximum value of $f(T)$ is $F_{10} = 89$.