I paint the board as the "supper evens" coloration (attachment 1). The piece will always cover exactly one blue square. Then there can be placed at most $2*4=8$ pieces. Now, lets paint the board as the "supper odds" coloration (attachment 2). The piece will always cover exactly one blue square. Because there are at most $8$ pieces one of the $9$ blue squares can not be covered. Leading to the total of covered squares being at most $24$. There is an example (attachments 3 and 4) and we are done . Aops wont let me put the 4th attachment so I let it to readers imagination

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$\color{green} \boxed{\textbf{SOLUTION}}$ It's easy to show that - It's possible to cover $24$ squares, FTSOC, We can cover $25$ squares, take two $5\times 5$ grids and color them like chess board such that $12$ squares are Black and $13$ squares are White for each grid. Consider the $4$ corner White squares of inner $3\times 3$ grid. It's easy to see that no matter wherever we place the $Z$ box It will go through one of it. So, we can place atmost $8$ $Z$ boxes If, We have to cover all $25$ squares we need to cover $13$ White squares. But for each case we can cover $8$ White squares and minimum $4$ Whites will be covered in both grid, giving maximum of $8+8-4=12$ White square covered. Contradiction $\blacksquare$