Inspired by Chung Kevin

If the centre square is blue, there have to be at least four white squares in the yellow region, and at least another four in the green region. Thus there would have to be at least eight white squares, which means that the number of blue squares could not exceed $17$ .

If the centre square is white, then there have to be at least three white squares in each of the yellow and green regions. This means that there have to be at least seven white squares, and hence at most $18$ blue squares. The second diagram shows that $\boxed{18}$ blue squares are possible.

I will show that the maximum is 18. FIrst, the diagram below shows that 18 is possible

Now, we will show that we cannot select 19 or more squares. Suppose not.
Let $a$ , $b$ , $\dots$ , $i$ denote the number of blue squares in the regions shown below.

Then $\begin{aligned} a + e + h + i &\le 5, \\ e + b + i + f &\le 5, \\ h + i + d + g &\le 5, \\ i + f + g + c &\le 5, \end{aligned}$ so $a + b + c + d + 2e + 2f + 2g + 2h + 4i \le 20.$

Suppose there were at least 19 blue squares, so $19 \le a + b + c + d + e + f + g + h + i.$ Then from the two inequalities above, $e + f + g + h + 3i \le 1.$ This tells us $i = 0$ , so $e + f + g + h \le 1$ . But $a$ , $b$ , $c$ , and $d$ are all at most 4, so $a + b + c + d + e + f + g + h + i \le 4 + 4 + 4 + 4 + 1 + 0 = 17,$ contradiction. Therefore, 18 is the maximum.