Three, in a certain city, there are $n$ high schools, the $i$-th high school sends $C_{i}$ students $\left(1 \leqslant C_{i} \leqslant 39,1 \leqslant i \leqslant n\right)$ to the gymnasium to watch a ball game, 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 does the gymnasium need to arrange at least to ensure that all students can be seated?
Three, the conclusion is that at least 12 rows are needed to ensure all students can be seated. Number the schools, and let the students from School No. 1 sit in the first row, followed by the students from School No. 2, and so on. In this way, students from each school sit together. If a row is full but there are still students from the same school, those remaining students will sit at the beginning of the next row. Since each row has 199 seats and the total number of students is 1990, after 10 rows, all students will be seated. Then, the students at the breaks can be moved, i.e., if the end of the first row and the beginning of the second row are occupied by students from the same school, all students from that school will be moved, and so on. In this way, at most 9 schools' students will be moved. Since each school has at most 39 students, and $5 \times 39 = 195 < 199$, the first 5 schools' students can be seated in the 11th row, and the remaining 4 schools' students in the 12th row. Thus, it is proven that arranging 12 rows can ensure all students are seated and meet the requirement: students from the same school sit in the same row.

If only 11 rows are arranged, it may indeed not be enough in some cases. For example, suppose $n=59$, with 58 schools each sending 34 students and 1 school sending 18 students, then $58 \times 34 + 18 = 1990$. Since $5 \times 34 = 170 < 199 < 204 = 6 \times 34$, except for one row that can seat 6 schools' students, the other 10 rows can each only seat 5 schools' students. Therefore, 11 rows can seat at most $6 + 5 \times 10 = 56$ schools' students, leaving two schools' students without seats. This shows that 12 rows are still the minimum number of rows needed to ensure all students can be seated according to the rules.