The problem describes a situation where we need to count how many different validated tickets can be created if each ticket has two punched holes, and the ticket can be rotated.

Step 1: Total possible punch combinations

A ticket has 9 possible locations where a punch can be made, and we need to select 2 different locations to punch. The number of ways to choose 2 locations out of 9 is calculated using the binomial coefficient:

$\binom{9}{2} = \frac{9 \times 8}{2} = 36$

Step 2: Accounting for symmetry (rotation)

The ticket can be inserted into the punching machine in either of two orientations (with the red arrow pointing up or down). Since the officials do not care about the orientation, any ticket configuration that can be obtained by rotating another configuration should be counted as the same.

For each pair of holes, if the locations are symmetric under a 180° rotation (i.e., the same pair of holes appear when the ticket is flipped), then it will only be counted once. If the pair is not symmetric, the two orientations will look different and should be considered distinct.

Step 3: Distinguishing symmetric from non-symmetric pairs

• A pair of punched holes is symmetric if and only if the holes are directly opposite each other when the ticket is rotated by 180°. In other words, they lie along a line of symmetry.
• For 9 possible locations, consider how many pairs of holes are symmetric.

The 9 positions can be visualized as the positions of a regular polygon (a nonagon). A symmetric pair would correspond to positions that are directly across from each other when flipped.

• After eliminating the symmetric cases, the number of distinct tickets can be calculated accordingly.

Thus, the number of different validated tickets the fee evader needs to collect is:

$\boxed{36}$