3. A city has $n$ high schools, the $i$-th high school sends $C_{i}\left(1 \leqslant C_{i} \leqslant 39,1 \leqslant i \leqslant n\right)$ students to the stadium to watch a ball game, the total number of all students is $\sum_{i=1}^{n} C_{i}=1990$, each row in the stand has 199 seats, and it is required that students from the same school must sit in the same row. How many rows does the stadium need to arrange at least to ensure that all students can be seated?
Since $199=5 \cdot 39+4$, at least 5 schools' students can be seated in each row.
Now, all students are seated by school, with no empty seats, and a total of 10 rows are arranged. In this way, the situation where students from the same school are not seated together can only occur at the end of one row and the beginning of the next, and the number of schools whose students are separated does not exceed 9, spanning $1-2,2-3, \cdots, 9-10$. If we move all the students from these 9 schools to sit in the 11th and 12th rows, then a total of 12 rows can ensure that all students can be seated. Since $1990=56 \cdot 35+30,199=5 \cdot 35+24$, if there are 57 schools, with 56 schools each having 35 students and 1 school having 30 students, in this case, each row can accommodate at most 5 schools' students, so 11 rows cannot ensure that all students can be seated.
In summary, at least 12 rows must be arranged to ensure that all students can be seated.