Given 2011*2011 square board, and (no limit for allowed number of napkins) square napkins size of 52*52. Once the board is fully covered by bunch of napkins then : Write on each cells of the board the numbers which tell us how many napkins are placed over that cell. Count the number of cells that are sharing identical number writen on them, For any possible arrangements of placing napkins on the board, what could be the maximum of such counts?
Since 2011=52*39-17 I can tell that maximum is at least (1994=2011-17) 1994*1994 but I lack of combinatorial techniques that can tighten the bound. I guess it must be under 3991049
I would appreciate it better if you can hint me the walkthrough by solving it under simpler conditions, such as (9*9 board , 4*4 napkin)
I'm not sure I understand the question quite correctly .... but to get the maximum number of napkins covering a square, you have the napkins one row or one column shifted from each other. This means that you get a large area in the middle of the board where all squares are covered by a napkin whose very right top cell covers that square, as well as a napkin whose top row and second from right square covers that square, etc. In other words, for each of the $52 \cdot 52$ cells from a napkin, there is a napkin that covers the square on the board by that cell. So all these squares would be covered by $52 \cdot 52$ = 2704$ napkins.
How many squares are there like that? Well, it's all the squares on the board that are at least $51$ squares removed from the border of the board, i.e. a $(2011-51-51)^2= 1909^2$ area. So, there are $1909^2=3,644,281$ squares that are covered by $2704$ napkins each.
If you still want the simpler example: you put the first napkin in the bottom left corner of the board, the second one one columns shifted to the right, etc. until the bottom row of squares is covered. This will take $6$ napkins of $4 \times 4$ size on a $9 \times 9$ board. Now go up one row and do the same ... and keep doing this until the $36$-th napkin covers the very top right square.
Now note that the $3 \times 3$ area of squares in the middle of the board are covered by $16$ napkins each: one napkin corresponding to each of the $16$ cells of the napkin.
Do you see how I got those numbers? The maximum number of napkins covering a cell is exactly the size of a napkind, so that is $4 \cdot 4 =16$. And the area in the middle that is covered by that maximum number of possible napkins is all the squares that are at least $4-1=3$ squares removed from the border, and so one side of that area is $9-3-3=3$, meaning that $3 \cdot 3=9$ squares are covered by $16$ different napkins each.
Now, with the much smaller example, note that each square in the $4$ strips of $3$ squares immediately outside and alongside the $3 \cdot 3$ area in the middle will each be covered by $12$ napkins, and the strips of $3$ outside of that are each covered by $8$. In fact, it turns out that $16$ squares of the board will be covered by $4$ napkins each, so the maximum number of squares being covered by the same number of napkins is $16$, rather than $9$.
In fact, if the question is asking for the maximum number of squares with the same coverage number, no matter what that coverage number is, you can do better yet: now, just place the napkins right next to each other without overlap ... except that at the end you'll have to have overlap. So, with a $9 \times 9$ board and $4 \times 4$ napkins, you can get a $5 \cdot 5$ area in a corner where each square is covered by exactly $1$ napkin, and you end up with $2$ strips of $5$ squares along the side, and $1$ square in the other corner that are also covered by $1$ square, so that gives a total of $25+10+1=36$ square covered by $1$ napkin each.
For the larger board from your question, putting all napkins all next to each other means that you are left with an uncovered strip of $17$ at two of the suides, meaning that covering those gives you an 'overlap strip' that is $35$ squares wide. This leaves $(37\cdot 52)+17)^2+2\cdot 17 \cdot (37 \cdot 52 + 17) + 17^2=1941^2+34 \cdot 1941 + 289 = 3,833,714$ squares covered by exactly one napkin.