11.6. In some cells of a $200 \times 200$ square, there is one chip - either red or blue; the other cells are empty. A chip can see another if they are in the same row or column. It is known that each chip sees exactly five chips of the other color (and possibly some chips of its own color). Find the maximum possible number of chips in the cells.

(I. Bogdanov)
Answer: 3800 chips.

Solution: An example containing 3800 chips can be constructed as follows. Highlight the "border" of width 5 in a $200 \times 200$ square. This border consists of four corner squares $5 \times 5$ and four rectangles $5 \times 190$. Place the chips in these four rectangles: red chips in the left and top rectangles, and blue chips in the right and bottom rectangles. It is easy to see that this example meets all the requirements and contains 1900 red and 1900 blue chips.

It remains to prove that there cannot be more than 3800 chips. Consider an arbitrary arrangement of chips that meets the requirements. Call a row (a row or a column) multicolored if it contains chips of both colors.

Let's make two useful observations. First, each chip sees some chip of the other color, so each chip lies in at least one multicolored row. Moreover, since a multicolored row contains a red chip, it cannot have more than five blue chips (otherwise, the red chip would see all of them). Similarly, a multicolored row cannot have more than five red chips, so it contains no more than 10 chips in total.

Now it is not difficult to obtain the required estimate. If there are 191 multicolored rows, then they contain no more than $191 \cdot 10 = 1910$ chips, and the remaining nine rows contain no more than $9 \cdot 200 = 1800$ chips, totaling no more than $1910 + 1800 = 3710$ chips. Then in $b$ rows, there are at most 10 chips each; on the other hand, each chip must lie in a multicolored row, so in the intersection of $n-a$ monochromatic columns and $n-b$ monochromatic rows, chips cannot be placed. Therefore, in $n-b$ monochromatic rows, there are no more than $a (\leqslant b)$ chips each, and the total number of chips is no more than $10b + (n-b)b = (n+10-b)b \leqslant 20(n-10)$ (the last inequality holds because $(n+10-b) + b = 20 + (n-10)$ and $n+10-b \leqslant 20 \leqslant n-10 \leqslant b)$.