Packed checkers

Use the pigeonhole principle: Divide the grid in four $4\times4$ squares: At least one square must have 12 blocks if there is 45 or more black squares.

We will prove first that in each subsquare just $12$ squares fit: Suppose $13$ fit. Let $c,e,b$ denonte the number of the squares in the corner, edge and "body" of the grid. Also, denote $v_{i,j}$ denote the number of "uses" of the cell $i,j$ and $n$ the total number of "uses", Here, we let a particular cell "use" each neighbor one time and use itself two times. So we can see that each cell has at most $4$ two uses, whichever cell we are talking about. So each corner contributes to $4$ in the total count, each edge $5$ , and each body $6$ , and in our $4\times4$ mini square we have $4c+5e+6b=n\le4\times16=64$ (four uses available for each one of the $16$ squares), and we want to maximize $c+e+b$ . Greedily using the cheapest first until exhaustion, we try $c=4,e=8,b=1$ . But in our diagram we see that is impossible. We can easily see that we can't exchange more than two of the border(corner or edge) squares with the body ones(that would cause that the total sum exceeds $64$ ). The only way of exchaning two borders without breaking our limit is removing two edges and adding two bodies, say: $c=4,e=6,b=3$ . But in the diagram, that is impossible. So we may exchange at most one square in the border with the one in the body. Back in our diagram, we see that that never works, since removing any single square in the border never lets us put two in the body.

So at most $12$ squares fit in one subsquare. But, can $12$ fit? In wich ways? First if we consider using just the borders we have a perfect example. A little juggling lefts us with the possibilities: grids

By simmetry and convinience, we can fix the subsquare with the missing top-right corner in the bottom left(this is the least we can influence the rest of the squares, it would be have the same effect if we chose the one with the top-right and bottom-left corners missing). We see that the possible squares(colored blue) for the subsquare in the right would be:

usables

Now, the minimal sum of neighbors using $11$ squares (with the lonely corner as contributing with three) is $3+4c+5e+6b=3+4\times2+5\times6+6\times2=53>52$

Since we had $13\times4$ uses at most in total (four for each square, 13 possible squares) this is impossible. So the $4\times4$ subsquare in the bottom right can have at most $10$ checkers and analogously the top-left subsquare must also have at most $10$ checkers. So in order to have at least $45$ total squares, the top-right subsquare must have at least $13$ checkers. Contradiction.

Imgur

I made a better picture.

Exactly my method. :D