3. Let $A$ be a set composed of several diagonals of a convex $n$-gon. If a diagonal in set $A$ intersects with exactly one other diagonal inside the convex $n$-gon, then it is called a "good" diagonal. Find the maximum possible number of good diagonals.
3. The maximum possible number of good diagonals is $2\left[\frac{n}{2}\right]-2$ (where $[x]$ denotes the greatest integer not exceeding the real number $x$).
Using induction, it is easy to prove a lemma.
Lemma: In a convex $n$-gon, at most $n-3$ diagonals can exist such that no two intersect internally.
The proof is left to the reader.
When $n=2,3,4$, the conclusion is obviously true.
Now assume $n \geqslant 5$.
Let $A_{1} A_{2} \cdots A_{n}$ be a convex $n$-gon, and $A$ be a set of some of its diagonals.

If any two good diagonals in set $A$ do not intersect internally, then by the lemma, the number of diagonals is at most $n-3$.

Now assume that there are two good diagonals in set $A$ that intersect internally, $A_{i} A_{k}$ and $A_{j} A_{l}$ (where $i<j<k<l$). Removing the quadrilateral $A_{i} A_{j} A_{k} A_{l}$ from the original polygon results in four polygons (which may degenerate into line segments)
$$
\begin{array}{l}
Q_{1}=A_{i} A_{i+1} \cdots A_{j}, Q_{2}=A_{j} A_{j+1} \cdots A_{k}, \\
Q_{3}=A_{k} A_{k+1} \cdots A_{l}, Q_{4}=A_{l} A_{l+1} \cdots A_{n} A_{1} A_{2} \cdots A_{i} .
\end{array}
$$

Apart from $A_{i} A_{k}$ and $A_{j} A_{l}$, the other good diagonals remain good diagonals of some polygon $Q$.
By the induction hypothesis, the number of good diagonals does not exceed
$$
\begin{array}{l}
2+(j-i-1)+(k-j-1)+(l-k-1)+(n-l+i-1) \\
=n-2 \text {. }
\end{array}
$$

When $n$ is odd, since the sum of the number of vertices of the polygons $Q$ is $n+4$, which is odd, at least one of the polygons $Q$ has an odd number of vertices. Therefore, in the above equation (1), one of the parentheses can still be reduced by 1, resulting in the estimate $n-3$.
An example where the equality is achieved is:
$$
A=\left\{A_{i} A_{n-i}, A_{i+1} A_{n-i+1}, i=1,2, \cdots,\left[\frac{n}{2}\right]-1\right\} .
$$

Each diagonal in set $A$ is good.