Given are sheets and the numbers $00, 01, \ldots, 99$ are written on them. We must put them in boxes $000, 001, \ldots, 999$ so that the number on the sheet is the number on the box with one digit erased. What is the minimum number of boxes we need in order to put all the sheets?
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   - The solution suggests using $34$ boxes.
   - The boxes listed are: $010, 101, 202, 303, 404, 505, 606, 707, 808, 909, 123, 321, 456, 654, 789, 987, 148, 841, 159, 951, 167, 761, 247, 742, 269, 962, 258, 852, 349, 943, 357, 753, 368, 863$.

3. **Verification:**
   - We need to verify that each number from $00$ to $99$ can be formed by erasing one digit from one of these $34$ boxes.
   - For example:
     - From $010$, we can get $01$ and $10$.
     - From $101$, we can get $10$ and $01$.
     - From $202$, we can get $20$ and $02$.
     - And so on.

4. **Generalization:**
   - Each box of the form $aba$ (where $a$ and $b$ are digits) can generate two distinct two-digit numbers: $ab$ and $ba$.
   - We need to ensure that all $100$ two-digit numbers are covered.

5. **Counting the Boxes:**
   - The provided solution uses $34$ boxes.
   - We need to check if fewer boxes can cover all $100$ two-digit numbers.

6. **Conclusion:**
   - The provided solution with $34$ boxes seems to cover all $100$ two-digit numbers.
   - To prove that $34$ is the minimum, we would need to show that no fewer boxes can cover all $100$ numbers, which involves a more detailed combinatorial argument.

The final answer is $\boxed{34}$.