Answer: 1 and all positive even numbers.

In the following, to operate on the $(i, j)$th cell (the cell in the $i$th row and the $j$th column) means to perform an operation by selecting the $(i, j)$th cell in the operation.

* When $n = 1$, it is clear that the goal can always be met.

* When $n$ is even, it suffices to show that it is possible to flip any arbitrary coin only in some moves. Indeed, if we want to flip the coin in the $(i, j)$th cell (called $C$), we simply operate on every cell that lies in the same row or the same column as $C$.

- The coin in $C$ is flipped $2n - 1$ times, which is odd.
- Each coin in the same row or the same column as $C$ other than the coin in $C$ is flipped $n$ times, which is even.
- Every other coin is flipped $2$ times, which is even.

Therefore, we see that only the coin in $C$ is flipped after these operations.

* When $n$ is odd and greater than $1$, it suffices to consider the reverse process, i.e. starting with all coins with the heads facing upwards, we cannot obtain all possible configurations after finitely many moves. Note that there are $2^{n^2}$ possible configurations since there are two possibilities for each of the $n^2$ cells. On the other hand, note that the order of operations does not affect the result, and operating the same cell twice does not change the result. Therefore, we may assume each cell is operated $0$ or $1$ time. This leads to $2^{n^2}$ different sets of operations. Thus, it suffices to show that there are two different sets of operations that lead to the same configuration.

Indeed, we simply operate on every cell in the first row and the first column except the $(1, 1)$th cell.
- The coin in the $(1, 1)$th cell is flipped $2(n-1)$ times, which is even.
- Each coin in the first row or the first column other than the coin in the $(1, 1)$th cell is flipped $n-1$ times, which is even.
- Every other coin is flipped $2$ times, which is even.

Therefore, we see that no coin is flipped after these operations, which means this has the same effect as not performing any operation at all (which is also one of the $2^{n^2}$ sets of operations). This completes the proof.