7. There is a checkered board of size $2 n \times 2 n$. Petya placed $(n+1)^{2}$ chips on it. The cat can, with one swipe of its paw, knock any one chip or two chips standing in adjacent side-by-side or corner-by-corner cells onto the floor. What is the minimum number of swipes the cat can definitely use to knock all the chips placed by Petya onto the floor? (S. Berlov, N. Vlasova)
Answer. $n^{2}+n$. Solution. Estimation. Divide the board into $n^{2}$ squares $2 \times 2$. The cat can sweep away any two chips that are in the same square. As long as there are more chips on the board than $n^{2}$, the cat has the opportunity to sweep away two chips, meaning he can do this at least $n+1$ times (after $n$ such actions, there are still $n^{2}+1>n^{2}$ chips on the board). By sweeping away the remaining chips one by one, the cat will need $(n+1)+\left(n^{2}-1\right)=n^{2}+n$ swipes. Example. Divide the board into $2 \times 2$ squares. Consider the diagonal running from the bottom left corner to the top right corner. Place 4 chips in the bottom left square, and in the others that this diagonal intersects, place 3 chips each, such that the bottom left cell remains empty. In all squares above the diagonal, place one chip in the top left corner, and in all squares below, place one chip in the bottom right corner. This results in exactly $\left(n^{2}-n\right)+3 n+1=(n+1)^{2}$ chips. Since the cat cannot, with the given arrangement of chips, sweep away chips from different squares in one swipe, to sweep away all the diagonal chips, at least $2 n$ swipes are needed, and for the remaining chips, $n^{2}-n$ swipes, making a total of $n^{2}+n$ swipes.