55. Answer: when $n=48$.

For each arrow, consider the three-cell corner obtained by removing from a $2 \times 2$ square, centered at the end of the arrow, a $1 \times 1$ square, the diagonal of which is this arrow. Note that such corners do not intersect and are contained within a $12 \times 12$ square. Therefore, there are no more than $12^{2} / 3=48$ such corners.

An example of placing 48 arrows is shown in the figure.

![](https://cdn.mathpix.com/cropped/2024_05_06_eb71c5b96b040a1dc9f3g-3.jpg?height=392&width=397&top_left_y=1660&top_left_x=835)
55. In a $10 \times 10$ grid (the sides of the cells have a unit length), $n$ cells were chosen, and in each of them, one of the diagonals was drawn and an arrow was placed on this diagonal in one of two directions. It turned out that for any two arrows, either the end of one coincides with the beginning of the other, or the distance between their ends is at least 2. What is the largest $n$ for which this is possible?