HMMT November

Number the rows and columns of the grid from 0 to 2013, thereby assigning an ordered pair to each tile. We claim that a tile $(i, j)$ may be selected if and only if $i \equiv j \equiv 0(\bmod 3)$; call such a square good. First, let us show that this condition is sufficient. Observe that any such square $s$ is the corner of a canonical $4 \times 4$ square $S$ whose vertices are all good. Then the sides of $S$ partition the board into nine distinct regions. It's easy to see that all of them can be suitably tiled. Now we show that only good squares can be removed. Let $\omega$ be a non-real cube root of unity. In the tile with coordinates $(i, j)$, place the complex number $\omega^{i+j}$. Note that any $1 \times 3$ or $3 \times 1$ rectangle placed on the grid must cover three squares with sum $1+\omega+\omega^{2}=0$. Now, note that the sum of the numbers on the whole $2014 \times 2014$ grid, including the removed tile, is $$\sum_{k=0}^{2013} \sum_{l=0}^{2013} \omega^{k+l}=\left(\sum_{k=0}^{2013} \omega^{k}\right)^{2}$$ which can be simplified to 1 using the identity $1+\omega+\omega^{2}=0$. Therefore, it is necessary that $i+j \equiv 0$ $(\bmod 3)$. By placing the complex number $\omega^{i-j}$ instead of $\omega^{i+j}$, the same calculations show that $i-j \equiv 0$ $(\bmod 3)$ is necessary. This can only occur if $i \equiv j \equiv 0(\bmod 3)$. Hence the answer is exactly the set of good squares, of which there are $672^{2}=451584$.