We call the first player A, and the second player B. It's easy to see that for \(n=1\), A wins by coloring the middle cell green. So let us now assume that \(n>1\). We claim that B wins always. Suppose the first move of A is the cell \((i,j)\). Then B locates the furthest corner cell from \((i,j)\) (we can assume WLOG that this furthest corner cell is \((1,1)\)), and colors its diagonally adjacent cell. Next B pairs the cells like this: If A colors one of these cells Green, B colors the corresponding paired cell red. If A colors any other cell, B colors any arbitrary cell except \((1,1)\). Since A has the last move, B can force A to color the cell \((1,1)\). This way, B wins, because there is no connection between the cell \((1,1)\) and A's first move \((i,j)\). In fact, no chain of Green cells starting from \((1,1)\) can get out of the first row or first column. Suppose such a chain lands on the second row (WLOG, the case for column is the same), and the first cell on second row it comes to is \((2,k)\) for \(k> 2\). So all the cells from \((2,3)\) to \((2,k-1)\) are red. As a result, all the cells from \((1,2)\) to \((1,k-2)\) are Green. But since \((2,k)\) is Green, \((1,k-1)\) is red. So we have \((1,k-2)\) is Green; both \((1,k-1)\) and \((2,k-1)\) are red; and \((2,k)\) is Green. So the green cells are not connected, contradiction!