The number $123454321$ is written on a blackboard. Evan walks by and erases some (but not all) of the digits, and notices that the resulting number (when spaces are removed) is divisible by $9$. What is the fewest number of digits he could have erased?

[i]Ray Li[/i]
1. **Calculate the sum of the digits of the original number:**
   The number is \(123454321\). The sum of its digits is:
   \[
   1 + 2 + 3 + 4 + 5 + 4 + 3 + 2 + 1 = 25
   \]

2. **Determine the condition for divisibility by 9:**
   A number is divisible by 9 if the sum of its digits is divisible by 9. Therefore, we need to find a subset of digits whose sum is divisible by 9.

3. **Find the remainder when the sum is divided by 9:**
   \[
   25 \mod 9 = 7
   \]
   This means we need to remove digits whose sum is 7 to make the remaining sum divisible by 9.

4. **Identify the fewest digits to remove to achieve the required sum:**
   - We need to find the smallest number of digits that sum to 7.
   - The digits of the number \(123454321\) are \(1, 2, 3, 4, 5, 4, 3, 2, 1\).

5. **Check combinations of digits to find the smallest set that sums to 7:**
   - The digit 7 is not present in the number.
   - We can remove \(2\) and \(5\) (since \(2 + 5 = 7\)).
   - Alternatively, we can remove \(3\) and \(4\) (since \(3 + 4 = 7\)).

6. **Verify the solution:**
   - Removing \(2\) and \(5\) leaves the digits \(1, 3, 4, 4, 3, 2, 1\), which sum to:
     \[
     1 + 3 + 4 + 4 + 3 + 2 + 1 = 18
     \]
     Since \(18\) is divisible by 9, this solution is valid.
   - Removing \(3\) and \(4\) leaves the digits \(1, 2, 4, 5, 4, 2, 1\), which sum to:
     \[
     1 + 2 + 4 + 5 + 4 + 2 + 1 = 19
     \]
     Since \(19\) is not divisible by 9, this solution is not valid.

Therefore, the fewest number of digits Evan could have erased is 2.

The final answer is \(\boxed{2}\)