For a math tournament, each person is assigned an ID which consists of two uppercase letters followed by two digits. All IDs have the property that either the letters are the same, the digits are the same, or both the letters are the same and the digits are the same. Compute the number of possible IDs that the tournament can generate.
1. **Calculate the number of IDs where the letters are the same:**
   - There are 26 possible uppercase letters.
   - The letters must be the same, so there are \(26\) choices for the letters.
   - There are 100 possible combinations of two digits (from 00 to 99).
   - Therefore, the number of such IDs is:
     \[
     26 \times 100 = 2600
     \]

2. **Calculate the number of IDs where the digits are the same:**
   - There are 10 possible digits (from 0 to 9).
   - The digits must be the same, so there are \(10\) choices for the digits.
   - There are 676 possible combinations of two uppercase letters (from AA to ZZ).
   - Therefore, the number of such IDs is:
     \[
     676 \times 10 = 6760
     \]

3. **Calculate the number of IDs where both the letters and the digits are the same:**
   - There are 26 possible uppercase letters.
   - The letters must be the same, so there are \(26\) choices for the letters.
   - There are 10 possible digits.
   - The digits must be the same, so there are \(10\) choices for the digits.
   - Therefore, the number of such IDs is:
     \[
     26 \times 10 = 260
     \]

4. **Apply the principle of inclusion-exclusion to avoid double-counting:**
   - We need to add the number of IDs where the letters are the same and the number of IDs where the digits are the same, then subtract the number of IDs where both the letters and the digits are the same (since these have been counted twice).
   - Therefore, the total number of possible IDs is:
     \[
     2600 + 6760 - 260 = 9100
     \]

The final answer is \(\boxed{9100}\)