In a convex $n$-gon, several diagonals are drawn. Among these diagonals, a diagonal is called [i]good[/i] if it intersects exactly one other diagonal drawn (in the interior of the $n$-gon). Find the maximum number of good diagonals.
To find the maximum number of good diagonals in a convex $n$-gon, we will use induction and some combinatorial arguments. Let's denote $g(n)$ as the maximum number of good diagonals in an $n$-gon.

1. **Base Cases:**
   - For $n = 2$ (a line segment) and $n = 3$ (a triangle), there are no diagonals, so $g(2) = 0$ and $g(3) = 0$.
   - For $n = 4$ (a quadrilateral), we can draw both diagonals, and they intersect each other exactly once, so $g(4) = 2$.

2. **Inductive Hypothesis:**
   Assume that for all $m < k$, $g(m) = 2\left\lfloor \frac{m}{2} \right\rfloor - 2$.

3. **Inductive Step:**
   Consider a convex $k$-gon with vertices labeled $A_1, A_2, \ldots, A_k$. Draw the diagonals $A_1A_3$ and $A_2A_4$. These diagonals intersect each other exactly once, making them good diagonals. Removing the quadrilateral $A_1A_2A_3A_4$ leaves a $(k-2)$-gon $A_1A_4A_5\ldots A_k$.

   By the inductive hypothesis, the $(k-2)$-gon can have at most $g(k-2) = 2\left\lfloor \frac{k-2}{2} \right\rfloor - 2$ good diagonals. Adding the two good diagonals $A_1A_3$ and $A_2A_4$, we get:
   \[
   g(k) \geq g(k-2) + 2 = 2\left\lfloor \frac{k-2}{2} \right\rfloor - 2 + 2 = 2\left\lfloor \frac{k-2}{2} \right\rfloor = 2\left\lfloor \frac{k}{2} - 1 \right\rfloor = 2\left\lfloor \frac{k}{2} \right\rfloor - 2
   \]
   Thus, $g(k) \geq 2\left\lfloor \frac{k}{2} \right\rfloor - 2$.

4. **Upper Bound:**
   We need to show that $g(n) \leq 2\left\lfloor \frac{n}{2} \right\rfloor - 2$. For $n = 2, 3, 4$, we have $g(2) = 0$, $g(3) = 0$, and $g(4) = 2$, which match $2\left\lfloor \frac{n}{2} \right\rfloor - 2$.

   For $k \geq 5$, assume $g(m) \leq 2\left\lfloor \frac{m}{2} \right\rfloor - 2$ for all $m < k$. Consider the cases where two good diagonals intersect in the interior of the polygon. Let these diagonals be $\overline{A_aA_c}$ and $\overline{A_bA_d}$ for $1 \leq a < b < c < d \leq n$. No other diagonals can intersect these two, so all other diagonals must be within the sub-polygons formed by these diagonals.

   Let $w = b-a+1$, $x = c-b+1$, $y = d-c+1$, and $z = (k-d+1) + a$. Then, $w + x + y + z = k + 4$ and $w, x, y, z \geq 2$. Thus, $w, x, y, z \leq k-2$. By the inductive hypothesis:
   \[
   g(k) \leq g(w) + g(x) + g(y) + g(z) + 2 \leq f(w) + f(x) + f(y) + f(z) + 2
   \]
   \[
   = \left(2\left\lfloor \frac{w}{2} \right\rfloor - 2\right) + \left(2\left\lfloor \frac{x}{2} \right\rfloor - 2\right) + \left(2\left\lfloor \frac{y}{2} \right\rfloor - 2\right) + \left(2\left\lfloor \frac{z}{2} \right\rfloor - 2\right) + 2
   \]
   \[
   = 2\left(\left\lfloor \frac{w}{2} \right\rfloor + \left\lfloor \frac{x}{2} \right\rfloor + \left\lfloor \frac{y}{2} \right\rfloor + \left\lfloor \frac{z}{2} \right\rfloor\right) - 6 \leq 2\left\lfloor \frac{w+x+y+z}{2} \right\rfloor - 6
   \]
   \[
   = 2\left\lfloor \frac{k+4}{2} \right\rfloor - 6 = 2\left\lfloor \frac{k}{2} + 2 \right\rfloor - 6 = 2\left\lfloor \frac{k}{2} \right\rfloor - 2
   \]
   Thus, $g(k) \leq 2\left\lfloor \frac{k}{2} \right\rfloor - 2$.

5. **Conclusion:**
   Since $g(n) \geq 2\left\lfloor \frac{n}{2} \right\rfloor - 2$ and $g(n) \leq 2\left\lfloor \frac{n}{2} \right\rfloor - 2$, we have $g(n) = 2\left\lfloor \frac{n}{2} \right\rfloor - 2$ for all $n \geq 2$.

The final answer is $\boxed{2\left\lfloor \frac{n}{2} \right\rfloor - 2}$