Kei0923 wrote:

On $5\times 5$ squares, we cover the area with several S-Tetrominos (=Z-Tetrominos) along the square so that in every square, there are two or fewer tiles covering that (tiles can be overlap). Find the maximum possible number of squares covered by at least one tile.

$\color{blue}\boxed{\textbf{Answer: 24}}$
$\color{blue}\boxed{\textbf{Proof:}}$
$\color{blue}\rule{24cm}{0.3pt}$
$\text{Let us consider the following coloring:}$
$[asy] draw((0,0)--(5,0)--(5,5)--(0,5)--(0,0)); draw((1,0)--(1,5)); draw((2,0)--(2,5)); draw((3,0)--(3,5)); draw((4,0)--(4,5)); draw((0,1)--(5,1)); draw((0,2)--(5,2)); draw((0,3)--(5,3)); draw((0,4)--(5,4)); fill((1,1)--(2,1)--(2,2)--(1,2)--cycle, black); fill((1,3)--(2,3)--(2,4)--(1,4)--cycle, black); fill((3,3)--(4,3)--(4,4)--(3,4)--cycle, black); fill((3,1)--(4,1)--(4,2)--(3,2)--cycle, black); [/asy]$
$\text{Let's note that each tetromino covers exactly one black square, as there can be at most 2 tiles per square and there are 4 black squares}$ $\Rightarrow \text{Number of tiles}\le 2\times 4=8$ $\text{Let us consider the following coloring:}$
$[asy] draw((0,0)--(5,0)--(5,5)--(0,5)--(0,0)); draw((1,0)--(1,5)); draw((2,0)--(2,5)); draw((3,0)--(3,5)); draw((4,0)--(4,5)); draw((0,1)--(5,1)); draw((0,2)--(5,2)); draw((0,3)--(5,3)); draw((0,4)--(5,4)); fill((0,0)--(1,0)--(1,1)--(0,1)--cycle, blue); fill((0,2)--(0,3)--(1,3)--(1,2)--cycle, blue); fill((0,4)--(0,5)--(1,5)--(1,4)--cycle, blue); fill((2,0)--(3,0)--(3,1)--(2,1)--cycle, blue); fill((2,2)--(3,2)--(3,3)--(2,3)--cycle, blue); fill((2,4)--(3,4)--(3,5)--(2,5)--cycle, blue); fill((4,0)--(5,0)--(5,1)--(4,1)--cycle, blue); fill((4,2)--(5,2)--(5,3)--(4,3)--cycle, blue); fill((4,4)--(4,5)--(5,5)--(5,4)--cycle, blue); [/asy]$
$\text{Let's keep in mind that each tetromino covers exactly one blue square, since there are 9 blue squares and there are at most 8 tiles,}$ $\text{then at least 1 square remains uncovered}$ $\text{Number of squares covered by at least one tile}\le 24$ $\color{blue}\rule{24cm}{0.3pt}$
$\color{blue}\boxed{\textbf{Example:}}$
$\color{blue}\rule{24cm}{0.3pt}$
$\text{We fill the board with two layers:}$
$[asy] draw((0,0)--(5,0)--(5,5)--(0,5)--(0,0)); draw((1,0)--(1,1)--(0,1)); draw((0,3)--(1,3)--(1,2)--(2,2)--(2,0)); draw((2,2)--(2,4)--(1,4)--(1,5)); draw((5,0)--(4,0)--(4,1)--(3,1)--(3,5)); draw((4,5)--(4,4)--(5,4)); draw((3,3)--(4,3)--(4,2)--(5,2)); [/asy]$
$[asy] draw((0,0)--(5,0)--(5,5)--(0,5)--(0,0)); draw((0,1)--(1,1)--(1,2)--(3,2)--(3,1)--(2,1)--(2,0)); draw((3,2)--(5,2)); draw((5,1)--(4,1)--(4,0)); draw((1,5)--(1,4)--(0,4)); draw((0,3)--(2,3)--(2,4)--(3,4)--(3,5)); draw((2,3)--(4,3)--(4,4)--(5,4)); [/asy]$
$\text{Note that there are 24 squares that meet the order}$ $\color{blue}\rule{24cm}{0.3pt}$
$\color{green}\boxed{\textbf{Conclusion:}}$
$\color{green}\rule{24cm}{0.3pt}$
$\boxed{\textbf{There are at most 24 squares covered by at least one tile}}_\blacksquare$ $\color{green}\rule{24cm}{0.3pt}$