Answer: 50. Solution: We will show that by the time the fastest runner finishes, any two runners of different colors have met exactly twice, from which the answer 2$\cdot$5$\cdot$5 = 50 will follow.

Let $s$ (km) be the length of the track. Set $T=2 s / 12$ (hours). Since the speed of the fastest runner is less than 12 km/h, he will run the track twice in a time greater than $T$. We will show that all possible meetings of runners will occur before $T$ hours have passed since the start.

Take two runners of different colors. Their first meeting will occur when they have together run the length of the track, and this will happen in less than $s / 18 = T / 3 < T$ hours. When the faster of the two has run the entire track, the slower one, whose speed is more than 9/12 = 3/4 of the faster one's speed, will have already run more than $3 / 4$ of the track length and thus the faster one, turning around, will not catch up (for this he would have to run at least four times faster than the slower one). Therefore, the second meeting of these two will occur when both are running back. By this time, they will have together run a distance of $3 s$, and this will happen in less than $3 s / 18 = T$ hours, which completes the proof.
1. 10 runners start simultaneously: five in blue T-shirts from one end of the running track, and five in red T-shirts from the other. Their speeds are constant and different, with each runner's speed being greater than 9 km/h but less than 12 km/h. Upon reaching the end of the track, each runner immediately turns around and runs back, and upon returning to their starting point, they finish running. The coach marks a check in the notebook each time two runners in different colored T-shirts meet (face to face or one overtakes the other), and no more than two runners meet at the same point during the run. How many checks will the coach have marked by the time the fastest runner finishes running? (I. Rubanov)