Too Negative Boxes!

Each row, column and diagonal must have either 1 or 3 negative signs for the product to stay negative, so there must be at least 1 negative sign per row, and thus at least 3 negative signs in total. The maximum possible number of sign changes is then 6, which can be achieved in the following way:

$\large + \space \space + \space \space -$ $\large + \space \space - \space \space +$ $\large - \space \space + \space \space +$

Each row/column has to have a different pairing of sign changes, for otherwise one of the columns/rows would have 3 positive signs, and there cannot be a sign change of the centre square, for otherwise one of the diagonals would have 3 positive signs.

Afterword: For a 4 by 4 grid we can change a maximum of 12 squares, as follows:

$\large + \space \space \space - \space \space\space + \space \space \space +$ $\large + \space \space \space + \space \space \space + \space \space \space -$ $\large + \space \space \space + \space \space \space - \space \space \space +$ $\large - \space \space \space + \space \space \space + \space \space \space +$

Theory: For an $n \times n$ grid, $n \gt 2$ , we can change a maximum of $n^{2} - n$ negative signs to plus signs.