Example 1 Given that after 20 gymnasts perform, 9 judges respectively assign them ranks from 1 to 20. It is known that, among the 9 ranks each athlete receives, the maximum and minimum differ by at most 3. Now, the sums of the ranks received by each person are arranged as $c_{1} \leqslant c_{2} \leqslant \cdots \leqslant c_{20}$. Find the maximum value of $c_{1}$. (All-Soviet Union, 2nd)
Analysis and Solution First, note that the goal of the solution is $c_{1} \leqslant P$, which is equivalent to the existence of an $i$ such that $c_{i} \leqslant P$. Therefore, estimating the total score of any player can provide an estimate for $c_{1}$.

To minimize the score (rank), it is desirable to have more judges rank the player first. Thus, we can estimate the score of the player who is ranked first the most. One case is self-evident, i.e., if a player gets 9 first places, it is clear that $c_{1} \leqslant 9$. Furthermore, when the "first" places are more evenly distributed, we should use an overall estimate, examining the total scores of the players who received first places. Let there be $r$ players who are ranked first.
(1) $r=1$, then $c_{1}=9$.
(2) $r=2$, then because there are 9 first places, at least one player $A$ gets 5 first places. Since the scores a player receives from different judges do not differ by more than 3, the other 4 scores of $A$ do not exceed 4. Therefore, the total score of $A$ is no more than $5+4 \times 4=21$, so $c_{1} \leqslant 21$.
(3) $r=3$, consider the total score $S$ of all players. They received a total of 9 first places, and there are another 18 ranks. Each rank's score is at most 4. Therefore, $S \leqslant 9+9 \times 3+9 \times 4=72$, so $c_{1} \leqslant \frac{72}{3}=24$.
(4) $r=4$, similarly estimate the total score of these 4 players, we have $S \leqslant 90$, so $c_{1} \leqslant\left[\frac{90}{4}\right]=22$.
(5) $r \geqslant 5$, each of these $r$ players' scores does not exceed 4, thus there are at least $9 r \geqslant 5 \times 9=45$ ranks not higher than 4. However, the 1 to 4 ranks given by 9 judges are at most $9 \times 4=36$, which is a contradiction.

In summary, $c_{1} \leqslant 24$.
Finally, $c_{1}=24$ is possible. In fact, let $c_{1}=c_{2}=c_{3}=(1+1+1)+(3+3+3)+(4+4+4), c_{4}=(2+2+2+2+2)+(5+5+5+5), c_{5}=(2+2+2+2)+(5+5+5+5+5)$, and $c_{i}=i+i+\cdots+i (i=6,7,8, \cdots, 20)$. See the table below:
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline Player & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & $\cdots$ & 20 \\
\hline 1 & 1 & 3 & 4 & 2 & 5 & 6 & 7 & 8 & 9 & $\cdots$ & 20 \\
\hline 2 & 1 & 3 & 4 & 2 & 5 & 6 & 7 & 8 & 9 & $\cdots$ & 20 \\
\hline 3 & 1 & 3 & 4 & 2 & 5 & 6 & 7 & 8 & 9 & $\cdots$ & 20 \\
\hline 4 & 3 & 4 & 1 & 2 & 5 & 6 & 7 & 8 & 9 & $\cdots$ & 20 \\
\hline 5 & 3 & 4 & 1 & 2 & 5 & 6 & 7 & 8 & 9 & $\cdots$ & 20 \\
\hline 6 & 3 & 4 & 1 & 5 & 2 & 6 & 7 & 8 & 9 & $\cdots$ & 20 \\
\hline 7 & 4 & 1 & 3 & 5 & 2 & 6 & 7 & 8 & 9 & $\cdots$ & 20 \\
\hline 8 & 4 & 1 & 3 & 5 & 2 & 6 & 7 & 8 & 9 & $\cdots$ & 20 \\
\hline 9 & 4 & 1 & 3 & 5 & 2 & 6 & 7 & 8 & 9 & $\cdots$ & 20 \\
\hline Rank Sum & 24 & 24 & 24 & 30 & 33 & 54 & 63 & 72 & 81 & $\cdots$ & 180 \\
\hline
\end{tabular}

Therefore, the maximum value of $c_{1}$ is 24.