Evaluation. Since each problem was solved by someone, no student could have solved five problems. If someone solved exactly four problems, then no student could have solved both the fifth and sixth problems together. Thus, there are 1000 students who solved the fifth problem, and another 1000 who solved the sixth. In this case, we have no fewer than 2001 high school students. Finally, if each student solved no more than three problems, then the number of solvers is no less than $6000: 3=2000$.

Example: 500 students solved problems 1, 2, and 3; 500 solved problems 3, 4, and 5; 500 solved problems 1, 5, and 6, and another 500 solved problems 2, 4, and 6.

## Answer

2000.
Zhendarov R.G.

In the spring round of the 2000 Cities Tournament, senior high school students from country $N$ were given six problems. Each problem was solved by exactly 1000 students, but no two students solved all six problems together. What is the minimum possible number of senior high school students from country $N$ who participated in the spring round?