If there is a prime number on the board, the player loses by definition. If there is an even number on the board that is not a power of $2$, then the player can always reduce it by its odd divisor, leaving an odd number on the board. If the number on the board is odd and is reduced by its (odd) divisor $a$, i.e. a number of the form $ab$ is replaced by $a(b - 1)$, the resulting number is even and is not a power of the pair. Therefore, if the starting number is even and not a power of even, the player can always make a move guaranteeing that for his next move he will again receive a natural number of the same kind. Thus, an even number that is not a power of even is a winning position, and an odd number is a losing position.

It remains to analyze the cases when the board number is $2^m$ for a natural $m$. In this case, the reducer should be $2^k$ for naturally $k < m$. If $k < m - 1$, then the resulting position $2^k(2^{m-k} - 1)$ is even and not a power of even, so it would be a winning move for the adversary and therefore an unacceptable move. So, we need $k = m - 1$. Playing this way, one player will get the even powers of $2$ on the board and the other the odd ones. Given that $2$ is a losing position, we conclude that even powers of $2$ are winning positions and odd powers of $2$ are losing positions.

The finally suitable $n$ are the even numbers that are not odd powers of the pair. Among the given there are $900 : 2 - 2 = 448$ such (we excluded $128 = 2^7$ and $512 = 2^9$). $\Box$