Example 1 A city has $n$ high schools, the $i$-th high school sends $c_{i}$ students $\left(1 \leqslant c_{i} \leqslant 39\right)$ to watch a ball game at the gymnasium, where $\sum_{i=1}^{n} c_{i}=1990$. Each row in the stand has 199 seats, and it is required that students from the same school sit in the same row. Question: What is the minimum number of rows that need to be arranged so that all students can definitely be seated?
Analysis and Solution First, consider what the worst-case scenario is. The worst-case scenario refers to the situation where the number of empty seats in each row is maximized. Clearly, if the number of participants from each school varies, it is easier to arrange, as the remaining seats can be filled by students from schools with fewer participants. Therefore, a worse scenario is when each school sends a relatively "uniform" and larger number of participants. Thus, we can first assume that all schools send the same number of participants. Suppose each school sends $r$ people, we examine what value of $r$ would maximize the number of empty seats in each row. The estimation is listed as follows:
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline Number of participants $r$ from each school & 39 & 38 & 37 & 36 & 35 & 34 & 33 & 32 & 31 & 30 & 29 & $r \leqslant 28$ \\
\hline Number of schools that can be arranged in each row & 5 & 5 & 5 & 5 & 5 & 5 & 6 & 6 & 6 & 6 & 6 & $\ldots$ \\
\hline Number of empty seats $t$ in each row & 4 & 9 & 14 & 19 & 24 & 29 & 1 & 7 & 13 & 19 & 25 & $t \leqslant 28$ \\
\hline
\end{tabular}

From the table, we can see that when each school sends 34 participants, the number of empty seats in each row is maximized at 29. Note that $1990=34 \times 58+18$. Therefore, we first examine the scenario where 58 schools each send 34 participants, and one school sends 18 participants, to see how many rows are needed. In this case, each row can accommodate a maximum of 6 schools, and the row that accommodates the school sending 18 participants can only be one row. Thus, the minimum number of rows required is $1+\left[\frac{58-6}{5}\right]+1=12$.

Finally, we prove that 12 rows are sufficient for any scenario. The most natural arrangement method is as follows:
First, arrange the first row so that the number of empty seats $x_{1}$ in the first row is minimized. Then, arrange the second row: again, minimize the number of empty seats $x_{2}$. Clearly, $x_{1} \leqslant x_{2}$. Otherwise, if $x_{1}>x_{2}$, moving the students from the second row to the first row would reduce the number of empty seats in the first row, contradicting the minimality of $x_{1}$. Continue this process until the 11th row, making the number of empty seats $x_{11}$ in the 11th row the smallest. We prove that with such an arrangement, all remaining students can definitely be arranged in the 12th row. $\square$

In fact, suppose there are still $x$ people left unseated after arranging 12 rows, then $x>x_{12}$. Otherwise, if $x \leq x_{12}$, these $x$ people could be seated in the 12th row. Therefore, $x=1990-\sum_{i=1}^{12}\left(199-x_{i}\right) \geqslant x_{12}$, which implies $x_{1}+x_{2}+\cdots+x_{11}>398$, so $398<x_{1}+x_{2}+\cdots+x_{11} \leqslant 11 x_{11}$, thus $x_{11} \geqslant 37$. This indicates that in the 12th row, the number of participants from each school is at least 38 (no school's participants can sit in the $x_{11}$ empty seats of the previous row). Since each school sends at most 39 participants, the schools seated in the 12th row must have sent 38 or 39 participants. Noting that $5 \times 39<199<6 \times 38$, exactly 5 schools are seated in the 12th row. Suppose $k$ schools send 38 participants and $5-k$ schools send 39 participants $(0 \leqslant k \leqslant 5)$, then the number of empty seats in the 12th row is
$$199-k \times 38-(5-k) \times 39=199+k-5 \times 39=4+k \leqslant 9 \text {. }$$

But $x_{11} \geqslant 37$, moving the students from the 12th row to the 11th row would reduce the number of empty seats in the 11th row, contradicting the minimality of $x_{11}$.