6.46 A number of students from a certain district participate in a mathematics competition. Each student's score is an integer, and the total score is 8250. The scores of the top three students are $88, 85, 80$, and the lowest score is 30. No more than 3 students can have the same score. How many students at minimum have a score of 60 or higher (including the top three)?
[Solution] Except for the top three, the scores are all between $30-79$, and the total score is $8250-(88+85+80)=7997$. To minimize the number of passing students, the number of failing students should be maximized, so as to take up the most points in the total score. Since it is known that at most 3 students can get the same score, we can make the number of people who get $30,31, \cdots, 59$ points each 3. The total score of these people is
$$
3 \times(30+31+\cdots+59)=4005 \text {, }
$$

The remaining total score is $7997-4005=3992$, which should be allocated to those who score $60-80$ points. Assuming that in addition to one third place, two others score 80 points, and those who score $60-79$ points each have 3 people, the total score should be
$$
2 \times 80+3 \times(60+61+\cdots+79)=4330 \text {. }
$$

And $4330-3992=338$. To minimize the number of people, the number of people removed should be maximized, with a total score of 338. First, remove three 60 points, leaving 158 points, at which point only two more people can be removed, let's say two 79 points. In the end, at least 60 people score no less than 60 points (including the top three).