Answer: 10. Solution: Estimation. To prove that there are no fewer than 10 huge numbers, it is sufficient to prove that among any 30 consecutive numbers, there will be a huge one. Indeed, suppose this is not the case. Then consider the largest of these 30 numbers. On one side of it, all 15 nearest numbers will be part of this set of thirty, and therefore will be less than it. But then, according to the condition, the 15 nearest numbers on the other side of it will also be less than it, meaning it will be huge. Contradiction. Example. All numbers ending in 0 can be arranged in the following order: 300, 280, 260, 240, ..., 20, 290, 270, 250, ..., 10. After them, in a similar order, we can arrange the numbers ending in 1: 291, 271, ..., 11, 281, ..., 1, and then similarly arrange the numbers ending in 2, and so on until the cycle is complete. It is not difficult to verify that all conditions will be met, and the huge numbers will be only those from 291 to 300.
4. In the vertices of a regular 300-gon, numbers from 1 to 300 are placed once each in some order. It turns out that for each number a, among the 15 nearest numbers to it in the clockwise direction, there are as many numbers less than a as there are among the 15 nearest numbers to it in the counterclockwise direction. A number that is greater than all 30 nearest numbers to it is called huge. What is the smallest possible number of huge numbers? (S. Berlov)