Think of the trivial solution. You just need to choose 4 squares as centers of the crosses. So for $n=4$ it can be done. Now we observe that with each move we can change at most 5 squares, so by pigeonhole principle we need at least 4 moves. For every even n we choose even times the same square (to achieve not changing anything) and then we repeat the process for $n=4$. Thus we can achieve changing every square from white to black for every even number of moves. We need to prove that for odd n it is impossible. Consider the squares you chose as centers of the crosses for $n=4$ and call them $A, B, C, D$. Now observe that we can't change the color of two of these squares simultaneously. Let $a, b, c, d$ be the number of moves that change color the square $A, B, C, D$ respectively. Clearly $a, b, c, d$ are all odd and according to the previous observation we have $a+b+c+d=n$. Therefore n must be even which completes the proof. (Forgive my poor English)

Similar to above but it's a little bit different. Answer:$n\geq 4$ and even. Example for $n\geq 4$ even: Take the centers as $(row,column):(1,3),(2,1),(3,4),(4,2)$ and take center as $(1,1)$ for $(n-4)$ times. It's obvious that we cannot make all the squares black in two moves because we can change at most $5$ squares' colours in one move. In one move, we always change exactly two of the squares in $S=(1,2),(1,3),(2,1),(2,4),(3,1),(3,4),(4,2),(4,3)$ which are on the edges but not on the corners. Also in one move, we can change the colours of two squares in $S$ if they have a common vertex. $P_1,P_2,...,P_8\in S$ is an $8-gon$ and we write non-negative integers on its edges. $|P_iP_{i+1}|$ is equal to the number of moves between $P_i$ and $P_{i+1}$. We know that $|P_iP_{i-1}|+|P_iP_{i+1}|$ is an odd number. So $4$ of the edges should be odd, $4$ of them should be even so the sum is even which is equal to the total number of moves, as desired.