$6\times 6$

Since 6x = 5x, the only possible value of x is zero. This is the same as Peter Byers' solution, and Wieter Jacobs looked at the exact same entries.

The value of each corner is equal to the sum x. Take the biggest blue line. It goes through 2 corners and 4 other circles. The sum of those 4 other circles are x - x - x = -x. Take the 4 diagonals that go through those 4 circles (so not the main diagonal). All the points covering those 4 diagonals have a total sum of 4x. Now look at those points, exept for the first 4. They go through exactly 4 diagonals, without crossing, so their sum is also 4x. We can observe that the sum of our 4 chosen points is 0. Now -x=0 so that x=0.

Now this does not account for a complete solution. But given the problem as stated, imagine we had filled in numbers and gotten our maximum x. We could just scale every number in the red dots by a factor $k>1$ to get a new maximum of $kx>x$ for all positive x (if for some reason x were negative to begin with multiply every number by -1). So by contradiction there cannot be a maximum x that is not zero. Thus if there is in fact a maximum for x its only feasible value would be 0, which of course can be achieved by just pluggin in 0 in all red dots.

Imagine the circles were each filled in with black or white, such that no two adjacent circles are the same color. Then the sum of all the numbers in the black circles can be shown to equal both 5x and 6x, so it follows that x=0.

==> this could also be another solution... nothing mentioned in the question to lead to a single definite solution.

Claim: The only (and hence maximum) value for $x$ is 0 for all square grids with an even number of red circles.

Proof: Suppose the size of the square grid is $2n$ . If $n = 1$ then each corner must have value $x$ as they lie on their own blue line however the longer diagonal blue lines have sum $2x$ . So we must have $x = 2x$ and so $x = 0$ .

Suppose $n > 1$ is the smallest value for which we can fill in the grid with $x \neq 0$ . Now consider the central $2(n - 1)$ square and do the following: For each red circle on the boundary of this square, add to its value the value of the neighbouring red circles which lie outside this square. This gives a square of size $2(n-1)$ with the same value for $x$ . This contradicts the minimality of $n$ and so $x = 0$ for all values of $n$ .