A Transformation of a cross-shaped grid filled with 1s (Proof of impossibility?) Consider a cross-shaped grid of size 7 as it shows on the figure (compared to one of size 3). Each cell contains a 1. Le'ts define a transformation $\pi$ of the grid as follows: take any 3 sized sub-cross of the grid and multiply all the cells inside by $-1$.

How many $\pi$ transformations are required to transform a cross-shaped grid of size 2017 that contains a 1 in each cell into a grid that contains $-1$ in every cell?
Any ideas on how to proceed? I was trying to solve the particular case for 7 but even for that I found it quite hard.

First, the case for grid size 2017.
Consider a grid of size $n > 3$. Reusing your drawing, consider the cells colored in red and yellow for any of the four sides of the grid:

Let us number those colored cells starting from one red cell and ending to the other red cell with indexes $1, \ldots, \frac{n-1}{2}$, so that cell $1$ and $\frac{n-1}{2}$ are the red ones.
Now define $\pi_1, \ldots, \pi_{\frac{n-1}{2}}$ the required number of transformations applied on the cells $1, \ldots, \frac{n-1}{2}$ (with the center of the 3 sized sub-cross on the cell).
$\pi_1$ and $\pi_{\frac{n-1}{2}}$ must be odd, because the corner cells are reachable only from cells $1$ and $\frac{n-1}{2}$ respectively. Then $\pi_2$ and $\pi_{\frac{n-3}{2}}$ must be even, because e.g. the border cell reachable from cell $1$ and $2$ must total an odd number of transformations, thus $\pi_1+\pi_2$ must be odd and similarly on the other side. We can continue the process along the side alternating even and odd transformations.
There are $\frac{n-1}{2}-2 = \frac{n-5}{2}$ yellow cells between the two red cells. If that number is even and it is for $n=2017$ but not for $n=7$, we will end up with the two cells $\frac{n-1}{4}$ and $\frac{n+3}{4}$ with $\pi_{\frac{n-1}{4}}$ and $\pi_{\frac{n+3}{4}}$ both even or both odd and thus $\pi_{\frac{n-1}{4}} + \pi_{\frac{n+3}{4}}$ even, so that the corresponding border cell, reachable from those cells, cannot be changed to $-1$.
Regarding the case $n=7$, consider the cells colored as below:

and with the usual notation, define $\pi_r$ the number of transformations applied on the red cells, and similarly $\pi_y$ for the yellow cells, $\pi_{p1}$ to $\pi_{p4}$ for the pink cells (choose whatever order you like), $\pi_g$ for the green cell.
$\pi_r$ must be odd, then $\pi_y$ must be even, as said above. Then the only way to have the pink cell $1$ at $-1$ is to have both $\pi_{p1}$ and $\pi_g$ odd or even, and similarly for pink cells $2,3,4$, therefore all pink cells must be odd or even, but this makes impossible changing the yellow cells to $-1$.
Maybe with a little more effort this can be extended for any other odd $n > 3$ with $\frac{n-1}{2}$ odd.
