Problem 1336

1. Understanding the Problem:
We need to fill a $2015 \times 2015$ grid with integers $1, 2, \ldots, n$ such that:
- For all $1 \le i \le n-1$, the cells with $i$ and $i+1$ share an edge.
- Cells with $1$ and $n$ also share an edge.
- No other pair of numbers share an edge.
- If two cells with $i < j$ share a vertex, then $\min\{j-i, n+i-j\} = 2$.

2. Visualizing the Grid:
We can visualize the grid as a cycle where each number $i$ is connected to $i+1$ and $n$ is connected back to $1$. This forms a Hamiltonian cycle.

3. Extending Cells:
- If a cell is a corner, extend it to a square with side $\frac{3}{2}$.
- If a cell is a straight, extend it to a rectangle with sides $1$ and $2$.

4. Calculating the Area:
- For corners, the area is $\left(\frac{3}{2}\right)^2 = \frac{9}{4}$.
- For straights, the area is $1 \times 2 = 2$.

5. Intersection of Extended Cells:
- Two extended cells intersect if and only if they are straights and they are adjacent to the same corner. Their intersection is a square with side $\frac{1}{2}$, giving an area of $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$.

6. Formulating the Inequality:
Let $x$ be the number of corners and $y$ be the number of straights. The total area covered by the extended cells should be less than or equal to the area of the $2016 \times 2016$ grid (since each cell is extended):
$$\frac{9}{4}x + 2y - \frac{1}{4}x \leq 2016^2$$
Simplifying, we get:
$$\frac{8}{4}x + 2y \leq 2016^2$$
$$2x + 2y \leq 2016^2$$
$$x + y \leq \frac{2016^2}{2}$$

7. Conclusion:
The maximum possible value of $n$ is $\frac{2016^2}{2}$.

The final answer is $\boxed{\frac{2016^2}{2}}$.