Example 6 (1990 National High School Mathematics League Second Test) A city has $n$ high schools, the $i$-th high school sends $C_{i}$ students to watch a ball game in the gym $\left(1 \leqslant C_{i} \leqslant 39, i=1,2, \cdots, n\right)$, the total number of all students is $\sum_{i=1}^{n} C_{i}=1990$. Each row in the stands has 199 seats. It is required that students from the same school must sit in the same row. How many rows at least should the gym arrange to ensure that all students can be seated?
Thought analysis: Utilize $1 \leqslant C_{i} \leqslant 39$, to analyze the minimum number of students that can sit in each row or the maximum number of empty seats that can be left in each row, to determine the minimum number of rows required.

Solution: Since $1 \leqslant C_{i} \leqslant 39$, each row can seat at least 161 people. Therefore, 13 rows can seat at least $161 \times 13 = 2093$ people, which is more than enough to seat all 1990 students.

Next, let's see if 12 rows of seats can accommodate all the students. Note that there are $199 \times 12 = 2388$ seats in total. After seating 1990 people, there are 398 seats left. Therefore, if the number of empty seats in each row does not exceed 33, all students can be accommodated.

Reorder the students from the largest to the smallest school, so that $C_{1} \geqslant C_{2} \geqslant C_{3} \geqslant \cdots \geqslant C_{n}$. There exists a non-negative integer $m$ such that $C_{m} \geqslant 34$ and $C_{m+1} \leqslant 33$ (if $m=0$, it means all $C_{i}$ do not exceed 33). Let $m=5p+r(0 \leqslant r<5)$, and arrange the students from the first $5p$ schools in the first $p$ rows, with 5 schools per row, each row seating at least 170 people, with at most 29 empty seats. Then, arrange the remaining schools in the next rows, seating until the next school's students cannot all be seated, ensuring that the number of empty seats in each row does not exceed 32, up to the 11th row. The total number of empty seats in these 11 rows does not exceed $32 \times 11 = 352$, so at least 1837 people have been seated, leaving at most $1990 - 1837 = 153$ people unseated, who can certainly be seated in the 12th row. Therefore, 12 rows of seats are sufficient to seat all the students as required.

Finally, let's consider the case with only 11 rows of seats. There are only 199 empty seats, and to accommodate all the students, the average number of empty seats per row cannot exceed 19. Now, let $n=80$, with the first 79 schools each having 25 students, and the last school having 15 students, so $25 \times 79 + 15 = 1990$. Except for one row that can seat $25 \times 7 + 15 = 190$ people, the remaining 10 rows can seat at most 175 people each, so 11 rows can seat at most $190 + 175 \times 10 = 1940$ people. This example shows that 11 rows of seats are insufficient. Therefore, to accommodate 1990 students, at least 12 rows of seats are required.