A problem on the $n\times n$ square 
This problem is stated directly as following
Problem
A $n\times n$ square that satisfies the following conditions:


*

*$n\geq 4$

*In each unit square we write a number $x$, with $0\le x \le 1 $. 

*No matter how we divide the square into 2 rectangle, the sum of the numbers written in at least one of them does not exceed 1.


Find the maximum value of the sum of the numbers written in the $n\times n$ square.
So far, I have been able to point out that there exists a scheme so that the answer to the problem is up to 5. As in the following figure. I wonder if anyone can prove this

 A: As quantus14 pointed out in the comment, a valid case is having a 5-cross of connected squares with 1 in each of them. So, to find the maximum, we may ignore cases with a sum of less than 5. 
Consider the left edge and shift it to the right one column at a time.
If the sum of the entire rectangle is $> 1$, then eventually, there will be a point where the left rectangle has a sum $> 1$. Just before this happens, mark the line as $L$. IE Rectangle left of $L$ has sum $ \leq 1$. Rectangle left of $L$ plus one more column has sum $> 1$.
If the sum of the entire rectangle is $\leq 1$, we ignore this case. 
Do the same for the Right R line, Top T line, Bottom B line.   
If $L$ happens to the right of $R$, then we can have an overlapping split of the rectangles into 2 with sum $ \leq 1$, so the total sum is $ \leq 2$. We ignore this case.
If the $L$ and $R$ lines coincide, then we have a split of the rectangles into 2, each with sum $ \leq 1$, so the total sum is $ \leq 2$. We ignore this case.
If $L$ is at least 2 columns to the left of $R$, then take any split by one of these internal lines, and we get 2 rectangles with a sum $> 1$ by definition. So this is not possible.
This leaves us with the remaining case where $L$ and $R$ differ by exactly 1 column.
Similarly for $T$ and $B$, they differ by 1 row.   
Now, sum of rectangle is $\leq$ left of L rectangle + right of R rectangle + top of T rectangle + bottom of B rectangle + $ 1 \times 1 $ unit square bounded by these 4 lines $\leq 1 + 1 + 1 + 1 + 1 = 5$.
Since the upper bound of 5 can be achieved, hence this is the maximum.     
Notes:
1. This problem is valid for all rectangles of the form $ n \times m$ with $ n \geq 3, m \geq 3$. We did not use the assumption of squares anywhere. The requirement on the dimensions is to allow for the "L, R, and extra column".
2. There are multiple equality cases. All that we require are
1) L/R, T/B to differ by 1 row/column,
2) sum of each left of L etc rectangles to be exactly 1,
3) overlap of any 2 left of L etc rectangles to have sum 0, and
4) $1\times 1$ square has value 1.
Can you come up with a simple description of all equality cases? 
