6. A rectangle $3 \times 100$ consists of 300 squares $1 \times 1$. What is the maximum number of diagonals that can be drawn in the squares so that no two diagonals share a common endpoint? (In one square, two diagonals can be drawn without sharing endpoints. Common internal points are allowed.)
Answer: 200.

Example. Let's number the rows and columns containing the squares. In each square with both odd numbers, we will draw two diagonals.

Another example. In all cells of the first and third rows, we will draw parallel diagonals.

![](https://cdn.mathpix.com/cropped/2024_05_06_24ccb12ba8aeda640d35g-3.jpg?height=492&width=463&top_left_y=619&top_left_x=947)

Estimate. We will color the vertices in two colors: vertices from odd-numbered columns in white, and even-numbered columns in black (see figure). There will be 200 white vertices and 204 black vertices. Each diagonal has ends of different colors, and

![](https://cdn.mathpix.com/cropped/2024_05_06_24ccb12ba8aeda640d35g-3.jpg?height=288&width=485&top_left_y=1139&top_left_x=921)
therefore, it is impossible to draw more than 200 diagonals.

Criteria. Estimate without an example: 4 points. Example without an estimate: 2 points.