Find all integers $n$ for which there exists a table with $n$ rows, $2022$ columns, and integer entries, such that subtracting any two rows entry-wise leaves every remainder modulo $2022$.

[i]Proposed by Tony Wang[/i]
To solve the problem, we need to find all integers \( n \) for which there exists a table with \( n \) rows and \( 2022 \) columns, such that subtracting any two rows entry-wise leaves every remainder modulo \( 2022 \).

1. **Assume the existence of such a table:**
   Let the rows of the table be denoted by \( \mathbf{a}_1, \mathbf{a}_2, \ldots, \mathbf{a}_n \), where each \( \mathbf{a}_i \) is a vector of length \( 2022 \) with integer entries.

2. **Condition on the differences:**
   For any two rows \( \mathbf{a}_i \) and \( \mathbf{a}_j \), the difference \( \mathbf{a}_i - \mathbf{a}_j \) must have all entries congruent modulo \( 2022 \). This means that for any \( k \) and \( l \), the difference \( a_{ik} - a_{jk} \equiv a_{il} - a_{jl} \pmod{2022} \).

3. **Consider three rows:**
   Suppose we have three rows \( \mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3 \). Define the sums of the differences:
   \[
   x = \sum_{k=1}^{2022} (a_{3k} - a_{1k}),
   \]
   \[
   y = \sum_{k=1}^{2022} (a_{1k} - a_{2k}),
   \]
   \[
   z = \sum_{k=1}^{2022} (a_{2k} - a_{3k}).
   \]

4. **Modulo condition:**
   By the problem's condition, each of these sums \( x, y, z \) must be congruent to the same value modulo \( 2022 \). Let \( m = 1011 \). Then:
   \[
   x \equiv y \equiv z \equiv m \pmod{2022}.
   \]

5. **Sum of differences:**
   Notice that:
   \[
   x + y + z = \sum_{k=1}^{2022} (a_{3k} - a_{1k}) + \sum_{k=1}^{2022} (a_{1k} - a_{2k}) + \sum_{k=1}^{2022} (a_{2k} - a_{3k}) = 0.
   \]
   This is because each term \( a_{ik} \) cancels out.

6. **Contradiction:**
   However, if \( x \equiv y \equiv z \equiv m \pmod{2022} \), then:
   \[
   x + y + z \equiv m + m + m \equiv 3m \pmod{2022}.
   \]
   Since \( m = 1011 \), we have:
   \[
   3m = 3 \times 1011 = 3033.
   \]
   But \( 3033 \not\equiv 0 \pmod{2022} \). This is a contradiction.

7. **Conclusion:**
   Therefore, it is not possible to have three rows satisfying the given condition. Hence, the maximum number of rows \( n \) for which such a table can exist is \( n \leq 2 \).

The final answer is \( \boxed{ n \leq 2 } \).