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.