Four. (50 points) A city has $n$ high schools, the $i$-th high school sends $C_{i}$ students to watch a ball game at the gymnasium $\left(0 \leqslant C_{i} \leqslant 39\right.$, $i=1,2, \cdots, n)$, the total number of all students is $C_{1}+C_{2}+\cdots+C_{n}=1990$. Each row of seats in the stand has 199 seats, and it is required that students from the same school must sit in the same row. What is the minimum number of rows the gymnasium needs to arrange to ensure that all students can be seated according to the requirements?
Four, Solution: Since $C_{i} \leqslant 39$, each row can accommodate at least 161 people. Therefore, 13 rows can accommodate at least $161 \times 13 = 2093$ people, which is sufficient to accommodate all students. Below, we use the extremal principle to prove that 12 rows are enough.

Since $C_{1}, C_{2}, \cdots, C_{n}$ are finite, the sums obtained by adding some of the $C_{i}$ that do not exceed 199 are also finite. Select the sum that is closest to 199, denoted as $C_{i 1} + C_{i 2} + \cdots + C_{i k}$. Arrange the students from these $k$ schools in the first row. Then, perform a similar operation on the remaining $C_{i}$, arranging the schools with the sum closest to 199 in the second row, and so on, up to the 10th row. Let the number of empty seats in each row be denoted as $x_{i}$. It is easy to see that $\{x_{i}\}$ is an increasing sequence.

If $x_{10} \geqslant 33$, then the remaining unseated students $C_{i}$ 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 to be seated in the 11th row, which will seat at least 170 students. Thus, $x_{11} \leqslant 29 \leqslant x_{10}$, which contradicts the minimality of $x_{10}$.

If $x_{10} \leqslant 32$, then the total number of empty seats in the first 10 rows does not exceed 320. Therefore, the first 10 rows have arranged at least 1670 people, and the number of unseated students does not exceed 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, there can only be 199 empty seats. To arrange all students, the average number of empty seats per row cannot exceed 19. Suppose $n=80$, the first 79 schools each have 25 people, and the last school has 15 people, totaling 1990 people. Except for one row that can seat $25 \times 7 + 15 = 190$ people, the other 10 rows can seat at most $25 \times 7 = 175$ people. Therefore, 11 rows can seat at most 1940 people, which shows that 11 rows are insufficient. Thus, at least 12 rows are needed.