Consider an $n\times n$ unit-square board. The main diagonal of the board is the $n$ unit squares along the diagonal from the top left to the bottom right. We have an unlimited supply of tiles of this form:
[asy]
size(1.5cm);
draw((0,1)--(1,1)--(1,2)--(0,2)--(0,1)--(0,0)--(1,0)--(2,0)--(2,1)--(1,1)--(1,0));
[/asy]
The tiles may be rotated. We wish to place tiles on the board such that each tile covers exactly three unit squares, the tiles do not overlap, no unit square on the main diagonal is covered, and all other unit squares are covered exactly once. For which $n\geq 2$ is this possible?

[i]Proposed by Daniel Kohen[/i]
1. **Understanding the Problem:**
   We need to cover an \( n \times n \) board with L-shaped tiles (each covering 3 unit squares) such that:
   - No tile covers any part of the main diagonal.
   - All other squares are covered exactly once.
   - Tiles can be rotated.

2. **Initial Observations:**
   - The main diagonal has \( n \) squares.
   - The total number of squares on the board is \( n^2 \).
   - The number of squares to be covered by tiles is \( n^2 - n \).

3. **Divisibility Condition:**
   - Each tile covers 3 squares, so \( n^2 - n \) must be divisible by 3.
   - This implies \( n(n-1) \) must be divisible by 3.
   - Since \( n \) and \( n-1 \) are consecutive integers, one of them must be divisible by 3.

4. **Base Cases:**
   - For \( n = 1 \): The board is \( 1 \times 1 \), and the main diagonal is the entire board. No tiles are needed. This is trivially possible.
   - For \( n = 2 \): The board is \( 2 \times 2 \), and the main diagonal has 2 squares. The remaining 2 squares cannot be covered by a single L-tile. This is not possible.
   - For \( n = 3 \): The board is \( 3 \times 3 \), and the main diagonal has 3 squares. The remaining 6 squares can be covered by 2 L-tiles. This is possible.

5. **General Case:**
   - We need to show that if \( n \) is good, then \( n+6 \) is also good.
   - Consider a \( (n+6) \times (n+6) \) board. We can split it into an \( n \times n \) board and a \( 6 \times (n+6) \) rectangle.
   - The \( n \times n \) board can be tiled by assumption.
   - The \( 6 \times (n+6) \) rectangle can be tiled using \( 2 \times 3 \) rectangles, as any nonnegative integer other than 1 can be written in the form \( 2a + 3b \) for \( a, b \geq 0 \).

6. **Exclusion of Specific Cases:**
   - For \( n = 4 \): The board is \( 4 \times 4 \), and the main diagonal has 4 squares. The remaining 12 squares cannot be covered without overlapping the main diagonal. This is not possible.
   - For \( n = 6 \): The board is \( 6 \times 6 \), and the main diagonal has 6 squares. The remaining 30 squares cannot be covered without overlapping the main diagonal. This is not possible.

7. **Conclusion:**
   - The pattern \( n \equiv 0, 1 \pmod{3} \) holds, excluding \( n = 4 \) and \( n = 6 \).
   - Therefore, the possible values of \( n \) are \( n = 1, 3, 7, 9, 10, 12, 13, \ldots \).

The final answer is \( \boxed{ n \equiv 0, 1 \pmod{3} } \) and \( n \neq 4, 6 \).