$5 \cdot 71$ A city has $n$ high schools, the $i$-th high school sends $C_{i}$ students to the stadium to watch a ball game. It is known that $0 \leqslant C_{i} \leqslant 39, i=1,2, \cdots, n, C_{1}+C_{2}+\cdots+C_{n}=1990$, each row of seats in the stand 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 stadium arrange to ensure that all students can be seated according to the requirements?
[Solution 1] Since $C_{i} \leqslant 39$, each row can accommodate at least 161 people. Therefore, with 13 rows, at least $161 \times 13 = 2093$ people can be seated, which is certainly enough for all 1990 students.

Next, we use the extremal principle to prove that 12 rows are sufficient. Since $C_{1}$, $C_{2}, \cdots, C_{n}$ are finite, their finite sums that do not exceed 199 are also finite. Select the finite sum that is closest to 199, denoted as $C_{i_{1}} + C_{i_{2}} + \cdots + C_{i_{k}}$, and arrange the students from these $k$ schools in the first row. Then, for the remaining $C_{i}$, we perform the same discussion and select the finite sum that does not exceed 199 and is closest to 199, and arrange the corresponding schools' students in the second row. Continue this process until the tenth row, and let the number of empty seats in the $i$-th row be $x_{i}$, where $i=1, 2, \cdots, 12$.

If $x_{10} \geqslant 33$, then the number of students $C_{i}$ from the remaining unseated schools are all at least 34. If the number of remaining schools is no more than 4, then 11 rows are sufficient. If the number of remaining schools is at least 5, then we can arbitrarily select 5 schools and arrange their students in the 11th row. At this point, $x_{11} \leqslant 29 < x_{10}$, which contradicts the minimality of $x_{10}$. If $x_{10} \leqslant 32$, then the total number of empty seats in the first 10 rows, $x_{1} + x_{2} + \cdots + x_{10} \leqslant 10 x_{10} \leqslant 320$, meaning that at least 1670 people have been seated in the first 10 rows, and the number of unseated students is at most 320. Since each row can seat at least 161 people, 12 rows are sufficient.

Finally, consider the case with only 11 rows. In this case, only 199 empty seats are allowed. To accommodate all students, the average number of empty seats per row cannot exceed 19. Suppose $n=80$, the first 79 schools each send 25 students, and the last school sends 15 students, totaling 1990 students. However, when arranging seats, except for one row that can seat $25 \times 7 + 15 = 190$ people, the other 10 rows can each seat at most 7 schools' 175 students. Therefore, 11 rows can seat at most 1940 people. This shows that 11 rows are insufficient.
In summary, at least 12 rows are needed.

[Solution 2] We only need to prove that 12 rows of seats are sufficient to ensure that all students can be seated as required. An example where 11 rows are insufficient is given in Solution 1.

First, arrange the first row according to the following principle: Starting from the first school, arrange the students in the first row of seats. Then, the second school, the third school, and so on, and as long as there are still empty seats, let the next school's students sit down until there are no more empty seats. At this point, the last school's students have at least one person seated, and either the school's students have filled the row or there are still students without seats. Clearly, the total number of students from these schools is at least 199.

Next, starting from the next school, arrange the students in the 2nd to 9th rows according to the same principle. Thus, the total number of students in each row is at least 199. Therefore, the remaining students are at most 199 and can be seated in the tenth row. Finally, move all the students from the last school in each of the first 9 rows out, and the rest of the students will be seated in the first 10 rows as required. Since it is not a problem to arrange 5 schools' students in each row, the 9 schools' students who were moved out can be arranged in the last two rows, thus proving that 12 rows of seats are sufficient.