Let $n>2$ be an integer. Anna, Edda and Magni play a game on a hexagonal board tiled with regular hexagons, with $n$ tiles on each side. The figure shows a board with 5 tiles on each side. The central tile is marked.
[asy]unitsize(.25cm);
real s3=1.73205081;
pair[] points={(-4,4*s3),(-2,4*s3),(0,4*s3),(2,4*s3),(4,4*s3),(-5,3*s3), (-3,3*s3), (-1,3*s3), (1,3*s3), (3,3*s3), (5,3*s3), (-6,2*s3),(-4,2*s3), (-2,2*s3), (0,2*s3), (2,2*s3), (4,2*s3),(6,2*s3),(-7,s3), (-5,s3), (-3,s3), (-1,s3), (1,s3), (3,s3), (5,s3),(7,s3),(-8,0), (-6,0), (-4,0), (-2,0), (0,0), (2,0), (4,0), (6,0), (8,0),(-7,-s3),(-5,-s3), (-3,-s3), (-1,-s3), (1,-s3), (3,-s3), (5,-s3), (7,-s3), (-6,-2*s3), (-4,-2*s3), (-2,-2*s3), (0,-2*s3), (2,-2*s3), (4,-2*s3), (6,-2*s3),  (-5,-3*s3),  (-3,-3*s3), (-1,-3*s3), (1,-3*s3), (3,-3*s3), (5,-3*s3), (-4,-4*s3), (-2,-4*s3), (0,-4*s3), (2,-4*s3), (4,-4*s3)};

void draw_hexagon(pair p)
{
draw(shift(p)*scale(2/s3)*(dir(30)--dir(90)--dir(150)--dir(210)--dir(270)--dir(330)--dir(30)));
}
{for (int i=0;i<61;++i){draw_hexagon(points[i]);}}
label((0,0), "\Large $*$");
[/asy]
    The game begins with a stone on a tile in one corner of the board. Edda and Magni are on the same team, playing against Anna, and they win if the stone is on the central tile at the end of any player's turn. Anna, Edda and Magni take turns moving the stone: Anna begins, then Edda, then Magni, then Anna, and so on.

The rules for each player's turn are:
[list]
[*] Anna has to move the stone to an adjacent tile, in any direction.
[*] Edda has to move the stone straight by two tiles in any of the $6$ possible directions.
[*] Magni has a choice of passing his turn, or moving the stone straight by three tiles in any of the $6$ possible directions.
[/list]
Find all $n$ for which Edda and Magni have a winning strategy.

To determine for which values of \( n \) Edda and Magni have a winning strategy, we need to analyze the movement rules and the structure of the hexagonal board. 

1. **Understanding the Board and Movement Rules:**
   - The board is a hexagon with \( n \) tiles on each side.
   - Anna moves the stone to an adjacent tile.
   - Edda moves the stone straight by two tiles in any of the six possible directions.
   - Magni can either pass his turn or move the stone straight by three tiles in any of the six possible directions.

2. **Coloring Strategy:**
   - We use a coloring strategy to determine the winning and losing positions.
   - Color the tiles in a repeating pattern such that:
     - Tiles where Edda moves are marked as \( b \) (winning for Edda and Magni).
     - Tiles where Magni moves are marked as \( c \) (winning for Magni).
     - The central tile is marked as \( * \).

3. **Pattern Analysis:**
   - The pattern repeats every three tiles in each direction.
   - If \( n = 3k + 1 \), the central tile \( * \) falls into a \( c \) tile position.
   - If \( n \neq 3k + 1 \), the central tile \( * \) does not fall into a \( c \) tile position.

4. **Winning Strategy for \( n = 3k + 1 \):**
   - When \( n = 3k + 1 \), the central tile is a \( c \) tile.
   - Edda and Magni can always move the stone to a \( c \) tile, reducing the distance to the central tile.
   - Anna cannot move the stone to a \( c \) tile directly from a \( b \) tile.
   - Therefore, Edda and Magni can always ensure the stone moves closer to the central tile, eventually reaching it.

5. **Non-Winning Strategy for \( n \neq 3k + 1 \):**
   - When \( n \neq 3k + 1 \), the central tile is not a \( c \) tile.
   - Anna can always move the stone to a \( b \) tile, preventing Edda and Magni from moving it to a \( c \) tile.
   - Thus, Edda and Magni cannot ensure the stone reaches the central tile.

Conclusion:
Edda and Magni have a winning strategy if and only if \( n = 3k + 1 \).

The final answer is \( \boxed{ n = 3k + 1 } \).