What is the maximum number of diagonals of a regular $12$-gon which can be selected such that no two of the chosen diagonals are perpendicular?

Note: sides are not diagonals and diagonals which intersect outside the $12$-gon at right angles are still considered perpendicular.

[i]2018 CCA Math Bonanza Tiebreaker Round #1[/i]
1. **Understanding the problem**: We need to find the maximum number of diagonals in a regular 12-gon such that no two diagonals are perpendicular. 

2. **Counting the diagonals**: The total number of diagonals in an $n$-gon is given by the formula:
   \[
   \frac{n(n-3)}{2}
   \]
   For a 12-gon, this becomes:
   \[
   \frac{12(12-3)}{2} = \frac{12 \cdot 9}{2} = 54
   \]
   So, there are 54 diagonals in a regular 12-gon.

3. **Identifying perpendicular diagonals**: In a regular 12-gon, diagonals are perpendicular if they intersect at the center and form a right angle. This happens when the diagonals connect pairs of opposite vertices. 

4. **Counting perpendicular pairs**: Each vertex can be connected to 5 other vertices to form a diagonal. For each of these diagonals, there are 5 other diagonals that can be perpendicular to it. However, this counts each pair of perpendicular diagonals twice, so we need to divide by 2:
   \[
   \frac{12 \cdot 5}{2} = 30
   \]
   Thus, there are 30 pairs of perpendicular diagonals.

5. **Maximizing non-perpendicular diagonals**: To maximize the number of diagonals such that no two are perpendicular, we need to avoid these 30 pairs. Since each pair involves 2 diagonals, we can select at most:
   \[
   54 - 30 = 24
   \]
   diagonals such that no two are perpendicular.

6. **Verification**: We need to ensure that this count is correct and that no two selected diagonals are perpendicular. By carefully choosing diagonals and avoiding the 30 pairs, we can indeed select 24 diagonals.

The final answer is \(\boxed{24}\).