Two types of tiles, depicted on the figure below, are given. 
[img]https://wiki-images.artofproblemsolving.com//2/23/Izrezak.PNG[/img]

Find all positive integers $n$ such that an $n\times n$ board consisting of $n^2$ unit squares can be covered without gaps with these two types of tiles (rotations and reflections are allowed) so that no two tiles overlap and no part of any tile covers an area outside the $n\times n$ board. \\ 
[i]Proposed by Art Waeterschoot[/i]
To solve this problem, we need to determine the conditions under which an \( n \times n \) board can be completely covered by the given tiles without any gaps or overlaps. The two types of tiles are:

1. A tile that covers 3 unit squares in an "L" shape (let's call it the "L-tile").
2. A tile that covers 4 unit squares in a "T" shape (let's call it the "T-tile").

We will use a coloring argument to determine the necessary conditions for \( n \).

1. **Coloring Argument for the L-tile:**
   - Consider coloring the \( n \times n \) board in a checkerboard pattern, where each unit square is either black or white, and adjacent squares have different colors.
   - Each L-tile covers 3 unit squares, which means it will always cover 2 squares of one color and 1 square of the other color.
   - Therefore, the number of L-tiles used must be even, as each L-tile changes the parity of the number of squares of each color.

2. **Coloring Argument for the T-tile:**
   - Consider coloring the \( n \times n \) board such that each square \((i, j)\) is colored black if both \( i \) and \( j \) are odd, and white otherwise.
   - Each T-tile covers 4 unit squares, and it will always cover 1 black square and 3 white squares.
   - Therefore, the number of T-tiles used must be a multiple of 4, as each T-tile changes the parity of the number of black squares by 1.

3. **Combining the Conditions:**
   - For the board to be completely covered by the tiles, the total number of unit squares must be divisible by both 3 and 4.
   - Since the board has \( n^2 \) unit squares, \( n^2 \) must be divisible by both 3 and 4.
   - The least common multiple of 3 and 4 is 12, so \( n^2 \) must be divisible by 12.
   - Therefore, \( n \) must be divisible by \( \sqrt{12} = 2\sqrt{3} \). Since \( n \) must be an integer, \( n \) must be divisible by 4.

4. **Additional Consideration:**
   - The smallest \( n \) that satisfies \( n \) being divisible by 4 is \( n = 4 \).
   - However, we need to check if \( n = 4 \) is a valid solution by attempting to tile a \( 4 \times 4 \) board with the given tiles.
   - It turns out that a \( 4 \times 4 \) board cannot be completely covered by the given tiles without gaps or overlaps.

Therefore, the smallest \( n \) that works is \( n = 8 \), and in general, \( n \) must be a multiple of 4 and greater than 4.

The final answer is \( \boxed{ n } \) is divisible by 4 and greater than 4.