6. In a $2019 \times 2019$ grid, select several $2 \times 2$ "square blocks" such that different "square blocks" do not overlap in any grid lines. How many square blocks can be selected at most?
【Answer】The maximum value sought is $\frac{2019^{2}-1}{8}=509545$.
【Solution】Let $n$ be the side length of the grid, in this problem $n=2019$.
As shown in the figure, for any odd number $n$, divide this grid into two stair-shaped parts along a zigzag line composed of a group of grids, with the side lengths of the two stair-shaped parts being $n-1$ and $n-3$. In each stair-shaped part, select checkerboard squares at intervals, totaling
$$
1+2+3+\ldots+\frac{n-1}{2}=\frac{n^{2}-1}{8} \text { squares. (3 points) }
$$

Color the selected checkerboard squares gray, and examine the adjacent connections between the gray squares and the uncolored white squares. Suppose a total of $x$ checkerboard squares are selected, with $b$ of them having one side on the edge of the grid, and $c \leq 4$ of them in the corners. Since no gray checkerboard squares are adjacent, the number of adjacent connections between all gray squares and white squares is $8(x-b-c)+6 b+4 c=8 x-2 b-4 c$. (6 points)

For a white square inside the grid, it can be adjacent to at most two selected checkerboard squares. Otherwise, the nearby checkerboard squares must have a common side, so it can be adjacent to at most 2 gray squares. (9 points)
For white squares on the edge (including corners), they are generally adjacent to only one gray square. Otherwise, the square must be on the edge, and there must be two gray squares on the edge next to it. Call such a white square a special square. Below, we explain: For each special square, we can find one or two white squares inside to correspond to it, either a white square not adjacent to any gray square, or two white squares each only adjacent to one gray square, and such a correspondence is not repeated.
(1) If $\mathrm{X}$ is a special square, then among the $\mathrm{ABC}$ three squares, either B is not adjacent to any gray square, or $A$ and $B$, or $B$ and C are only adjacent to one gray square, at this time, $\mathrm{X}$ corresponds to these squares. (12 points)
(2) If the above correspondence produces a repetition, there must be special squares X and Y both adjacent to a corner checkerboard square, and both adjacent to another edge checkerboard square. At this time, if $\mathrm{A} 3$ is not occupied by a checkerboard square, then $\mathrm{A}, \mathrm{A} 1, \mathrm{~A} 2$ can correspond to $\mathrm{X}, \mathrm{Y}$ respectively.
(3) If $\mathrm{A} 3$ is occupied, examine the white squares $\mathrm{Z}$ and $\mathrm{W}$, if one of them is not a special square, let's say $\mathrm{Z}$, then B1, B2, B3 have a square that can correspond to $\mathrm{Y}$, and $\mathrm{A}$ corresponds to $\mathrm{X}$.
(4) If $\mathrm{Z}$ and $\mathrm{W}$ are both special squares, examine B5 and C5, if one of them is not occupied, let's say B5, then B2 can correspond to $\mathrm{Z}$, B3, B4 correspond to $\mathrm{Y}$, and $\mathrm{A}$ corresponds to $\mathrm{X}$.
(5) If $\mathrm{B} 5$ and $\mathrm{C}$ are both occupied, then $\mathrm{D}$ is not adjacent to any gray square, $\mathrm{D}$ can correspond to $\mathrm{Y}$, and $\mathrm{A}$ corresponds to $\mathrm{X}$. Because $n>11$, the squares D corresponding to each corner shape are different.

After explaining the above correspondence, the number of connections between all inner white squares and gray squares will not exceed $2 \times$ the number of all inner white squares $-2 \times$ the number of special squares. (15 points)

Consider the closed zigzag line formed by the edges of the square, it is $4 n$ long, gray squares occupy $2 b+4 c$ length, so white squares occupy $4 n-2 b-4 c$ length, thus the number of edge white squares is $4 n-2 b-4 c-(4-c)=4 n-2 b-3 c-4$.

And all white squares are divided into $b+c$ intervals by several checkerboard squares on this closed zigzag line. Therefore, the number of intervals of length 1 is at least $2(b+c)-(4 n-2 b-3 c-4)=4 b+5 c-4 n+4$, which is also the number of special squares, note that this number can be negative, but it does not affect the minimum property. (18 points)

Now we re-examine the number of connections between white squares and gray squares, inner white squares provide at most $2 \times$ the number of all inner white squares $-2 \times$ the number of special squares, edge white squares provide at most the number of edge white squares + the number of special squares, so the total is no more than
$2 \times$ the total number of white squares - the number of edge white squares - the number of special squares
$$
\begin{array}{l}
=2\left(n^{2}-4 x\right)-(4 n-2 b-3 c-4)-(4 b+5 c-4 n+4) \\
=2 n^{2}-8 x-2 b-2 c . \text { (21 points) }
\end{array}
$$

These two connection numbers are essentially the same, so $8 x-2 b-4 c \leq 2 n^{2}-8 x-2 b-2 c$.
Rearranging gives $16 x \leq 2 n^{2}+2 c, x \leq \frac{n^{2}+c}{8} \leq \frac{n^{2}+4}{8}$. When $n$ is odd, the maximum integer $x$ can take is $\frac{n^{2}-1}{8}$.