Answer: 1342. Solution: Example. We will color the second, third, fifth, sixth, ..., 2012th, and 2013th cells in the first vertical column of the board. Then, in all 3x3 squares adjacent to the left edge of the board, exactly two cells are colored, and in all other colored cells, there are none. In this case, a total of $2 \cdot 2013 / 3=1342$ cells are colored.

Estimate: Suppose fewer than 1342 cells are colored on the board. We will call a triplet of consecutive horizontal rows of the board weak if at least two of these rows do not contain any colored cells. We will show that weak triplets exist. Indeed, if we divide all horizontal rows of the board, except the first one, into triplets of consecutive rows, we get 671 triplets. If none of them are weak, then each triplet contains at least two colored cells, and a total of at least $671 \cdot 2=1342$ cells are colored - a contradiction.

Note that there will be a weak triplet that contains colored cells. Indeed, take any weak triplet. If it does not contain any colored cells, then by moving it towards one of the colored cells, we will eventually encounter a row with colored cells and obtain the desired triplet. Fix it, remove the three rightmost cells from the rectangle $2014 \times$, formed by its horizontal rows, and divide the remaining rectangle $2013 \times 3$ into 671 squares $3 \times 3$. In each of these squares, either two cells are colored or none. If each has two, then a total of at least 1342 cells are colored - a contradiction. Therefore, there will be a square without any colored cells. We will move it horizontally towards the nearest colored cell. When it first encounters a colored cell, we will get a square with exactly one colored cell - a contradiction.
7. On a grid board of size $2014 \times 2014$, several (no less than one) cells are colored such that in each $3 \times 3$ square, an even number of cells are colored. What is the smallest possible number of colored cells? (M. Antipov)