Mongolian Mathematical Olympiad

Answer: $n$ even.

Let's start by proving that any even number $n$ is a good number. For the $n=2k$ case we can construct a checkered pattern that all the cells of the frame contains checker and contains $2k-4$ case in the middle. The figure below illustrates this construction:

Now, let's show that for an odd number $n=2k+1$, each cell of the board cannot have exactly two checkered neighbors. To demonstrate this, we will focus on the diagonal cells of the table. Consider the neighbors of the cells $(i,i)$ for $i\in [n]$. Starting with cell $(1,1)$, since it has exactly two neighbors, we place checkers in those neighboring cells, which are $(1,2)$ and $(2,1)$. Now, these cells become neighbors of cell $(2,2)$, so we do not place checkers in cells $(2,3)$ and $(3,2)$. We continue this process for cells $(1+2j,2+2j)$ and $(2+2j,1+2j)$, where $j$ ranges from 0 to $k-2$. As a result, cells $(1+2j,2+2j)$ and $(2+2j,1+2j)$ have checkers, while cells $(2+2j,3+2j)$ and $(3+2j,2+2j)$ do not contain checkers. The figure below illustrates this construction: Now, consider cells $(2k,2k+1)$ and $(2k+1,2k)$. Since these cells are neighbors of the corner cell $(2k+1,2k+1)$, they should be checkered. However, we have already established that these cells cannot be checkered. This leads to a contradiction. Therefore, the number $n=2k+1$ is not a good number.

In conclusion, we have shown that an even number is a good number, while an odd number is not a good number.