Problem assumes that we remove $2(i-1)$ squares if $i\leq n$ , and $2(2n-i)$ squares if $i>n$ .
Divide the entire board into 4 quadrants each containing $n^2$ unit squares.
First we note that the $2$ squares on the center on each of the $4$ bordering lines of the board can always be completely covered by a single tile, so we can count in the first and last unit squares (which are diagonally opposite) of each quadrant as being filled in completely by a tile.
So in each quadrant we have:
if $n$ is even, there are exactly $(n-4)/2$ unit squares which cannot be filled by the tiles and if $n$ is odd, there are exactly $(n-3)/2$ unit squares which cannot be filled by the tiles.
Above can be seen by drawing a diagram and noticing that alternate columns have even and odd number of unit squares (leaving a unit square uncovered by tiles in odd numbered blocks of columns).
Also, note that the total number of unit squares which were removed from each quadrant = $(1 + 2+ 3 +... n-1) = n(n-1)/2$
Let us consider the 2 cases for parity of $n$ :
$Case 1: n$ is even
for $n = 2$ , it can be seen easily that we can use a maximum of $6$ tiles.
for $n \ge 4:$
Total number of squares that cannot be filled in each quadrant is: $n(n-1)/2 + (n-4)/2 = (n^2 - 4)/2$
So total number of squares that cannot be filled on the entire board = $2(n^2 - 4)$
So total number of squares that CAN be filled completely by the tiles = $4n^2 - 2(n^2 - 4) = 2n^2 + 8$
So the maximum number of tiles that can be used = $(2n^2 + 8)/2 = n^2 + 4$
$Case 2: n$ is odd
for $n = 1$ , it can be seen easily that we can use a maximum of $2$ tiles.
for $n \ge 3:$
Total number of squares that cannot be filled in each quadrant is: $n(n-1)/2 + (n-3)/2 = (n^2 - 3)/2$
So total number of squares that cannot be filled on the entire board = $2(n^2 - 3)$
So total number of squares that CAN be filled completely by the tiles = $4n^2 - 2(n^2 - 3) = 2n^2 + 6$
So the maximum number of tiles that can be used = $(2n^2 + 6)/2 = n^2 + 3$
$Kris17$