Problem 10.3. Petya and Daniil are playing the following game. Petya has 36 candies. He lays out these candies in the cells of a $3 \times 3$ square (some cells may remain empty). After this, Daniil chooses four cells forming a $2 \times 2$ square and takes all the candies from there. What is the maximum number of candies that Daniil can guarantee to take?
# Answer: 9.

Solution. If Petya places 9 candies in each corner cell (and does not place any candies in the other cells), then in any $2 \times 2$ square there will be exactly 9 candies. After this, Daniil will be able to take only 9 candies.

Let's prove that Daniil can get at least 9 candies. Suppose the opposite: let him be able to take no more than 8 candies for any choice of a $2 \times 2$ square. Let him then sequentially take candies from all $x$ four $2 \times 2$ squares: first from one (no more than 8 candies), then the remaining candies from the next square (also no more than 8), then from the third square, and then from the fourth. At each step, he could take no more than 8 candies, so in total he will take no more than 32. But there should be 36. Contradiction.