Say you have a column $C_i$ with negative sum, and you change the sign of all elements in that column. This must increase the total sum of all the elements in the grid (hereby just called the "total sum"), since all the other columns are unaffected, and the sum of all the column sums is the same as the total sum. The same thing goes for rows, of course.
There are only finitely many different total sums that can be made with the numbers on the board by changing the signs, even if you could freely change the signs of single elements. As long as there is a row or column with negative sum, you can increase the total sum, and increasing the total sum can be done only finitely many times. That means that the algorithm "Find a row or column with negative sum and flip the signs in it, then repeat" must end after a finite number of repeats.
Yes, this is true. Take $A$ to be an admissible set achieving the highest possible $m(A),$ but subject to that condition, with the minimum achievable number of cells equal to $m(A).$ (So we prefer one entry to equal $m(A)$ rather than two.)
Without loss of generality, $A$ is the set of diagonal cells, and the $(1,1)$ cell is $m(A).$ Consider which cells are not in $A$ but are strictly greater than $m(A).$ Call these "augmenting".
If there is a row with no augmenting cells, then the element of $A$ in that row is at least $1-2m(A),$ which gives $s(A)\geq 1.$ So we are done unless each row has an augmenting cell.
If the $(2,1)$ and $(3,1)$ cells are not augmenting, then the total of the cells in the $(2,2),(2,3),(3,2),(3,3)$ positions is at least $2-2m(A).$ Therefore either of $\{(1,1),(2,2),(3,3)\}$ or $\{(1,1),(2,3),(3,2)\}$ has sum at least $1,$ and by the previous paragraph these both have minimum at least $m(A),$ so we're done.
The remaining case is that one of the $(2,1)$ or $(3,1)$ cells is augmenting. Some cell in the first row is also augmenting. If $(1,2)$ and $(2,1)$ are augmenting then the set $(1,2),(2,1),(3,3)$ is an admissible set contradicting the choice of $A$ - it either has a higher minimum or the same minimum but fewer elements achieving the minimum. Similarly we are done if $(1,3)$ and $(3,1)$ are augmenting. Otherwise wlog $(2,1),(1,3)$ are augmenting and $(3,1),(1,2)$ are not (the other case just swaps $2$ and $3$). Then $(3,2)$ must be augmenting because it is the only remaining cell in the third row, so $(2,1),(1,3),(3,2)$ forms an admissible set with higher minimum.
Best Answer
Short answer: 8/7 based on the matrix $$\begin{pmatrix}0.5 & 0.5 & 0 \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3}\end{pmatrix}.$$ for which you get $X=4/3$ and $Y=7/6$ (by coloring the (1,1), (2,2) and (2,3) elements).
How I found this: consider the matrix $A\in\mathbb{R}_+^{2\times k}$ with elements $a_{ij}\geq 0$. We can always permute the columns in a way that the largest numbers are in the first row for the first set of columns, and in the second row for the last set of columns. Let $p$ denote the number of columns for which the maximum is in the first row. We will consider $p=1,2,\ldots,\left \lfloor{k/2}\right \rfloor$ (due to symmetry there is no need to consider larger $p$). We have: $$X=\sum_{j=1}^p A_{1j} + \sum_{j=p+1}^k A_{2j}.$$ To further reduce symmetry, we can assume that the first $p$ columns and the last $k-p$ columns of $A$ are both sorted high to low in the first row. This symmetry breaking rule is not necessary, but helps in terms of speed (the first one is necessary for efficiently setting $X$). Let $C \subset \{0,1\}^{2 \times k}$ denote the set of $2^k$ possible colorings: for $c\in C$, $c_{ij} = 1$ if element $(i,j)$ is colored, and 0 otherwise. The following inequalities set the highest lower bound on $Y$ by enumerating all valid colorings: $$\forall c \in C : (Y \geq \sum_{ij} a_{ij} c_{ij}) \vee \sum_{j} a_{1j} c_{1j} < a_{1j} (1-c_{1j}) - a_{1j'} \forall j' \vee \sum_{j} a_{2j} c_{2j} < a_{2j} (1-c_{2j}) - a_{2j'} \forall j',$$ where $j'$ has to be an uncolored element $(c_{ij'}=0)$.
For fixed $k$, $p$ and $X$, you can minimize $Y$ via mixed integer optimization (matlab/YALMIP code below). It uses a big-M formulation for the three alternatives above. Here is the output for $k=4$:
If you run the code with $k$ larger than 4, you just get extra columns that are zero. Zooming in (generating more data points) shows a maximum at $X=4/3$.