In $10\times 10$ square we choose $n$ cells. In every chosen cell we draw one arrow from the angle to opposite angle.  It is known, that for any two arrows, or the end of one of them coincides with the beginning of the other, or
the distance between their ends is at least 2. What is the maximum possible value of $n$?
To solve this problem, we need to understand the constraints and how they affect the placement of arrows in the $10 \times 10$ grid. Let's break down the problem step by step.

1. **Understanding the Arrow Placement:**
   Each arrow is drawn from one corner of a cell to the opposite corner. For example, an arrow in cell $(i, j)$ can go from $(i, j)$ to $(i+1, j+1)$ or from $(i, j+1)$ to $(i+1, j)$. 

2. **Constraints:**
   - 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.
   - This means that no two arrows can end within a distance of less than 2 units from each other unless they are directly connected.

3. **Grid Analysis:**
   Consider the $10 \times 10$ grid. Each cell can be represented as $(i, j)$ where $0 \leq i, j \leq 9$. 

4. **Distance Calculation:**
   The distance between the ends of two arrows is calculated using the Euclidean distance formula. For two points $(x_1, y_1)$ and $(x_2, y_2)$, the distance is:
   \[
   d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}
   \]
   For the distance to be at least 2, we need:
   \[
   \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \geq 2
   \]

5. **Maximizing $n$:**
   To maximize the number of arrows $n$, we need to place arrows such that the distance constraint is satisfied. One effective way to do this is to place arrows in a checkerboard pattern. In a checkerboard pattern, no two arrows will be within a distance of 2 units unless they are directly connected.

6. **Checkerboard Pattern:**
   In a $10 \times 10$ grid, a checkerboard pattern will have 50 cells of one color and 50 cells of the other color. We can place arrows in all cells of one color (say black) without violating the distance constraint.

7. **Conclusion:**
   By placing arrows in a checkerboard pattern, we can place arrows in 50 cells without any two arrows violating the distance constraint.

The final answer is $\boxed{50}$.