Example 3: On a lottery ticket, there are 50 spaces arranged in sequence. Each participant fills in the integers from 1 to 50 on their ticket (each number appears exactly once on the same ticket). The host also fills out a ticket as the reference. If a participant's number sequence has a position where the number matches the number in the corresponding position on the reference ticket, they win a prize. How many tickets does a participant need to fill out at least to ensure they will definitely win? (25th All-Russian Math...)
Analysis and Solution If filling $k$ lottery tickets can result in a win, then $k$ is called a winning number. Arrange the $k$ filled lottery tickets into a $k \times 50$ number table:
\begin{tabular}{|c|c|c|c|c|c|}
\hline & $a_{1}$, & $a_{2}$, & $a_{3}$, & $\cdots$, & $a_{50}$ \\
\hline & $b_{1}$, & $b_{2}$, & $b_{3}$ & * & $b_{50}$ \\
\hline & * & * & * & … & *. \\
\hline & $c_{1}$, & $c_{2}$, & $c_{3}$, & $\cdots$ & $c_{50}$ \\
\hline Bottom ticket & $x_{1}$, & $x_{2}$, & $x_{3}$, & , & $x_{50}$ \\
\hline
\end{tabular}

A number $k$ is called winning if, regardless of how $x_{1}, x_{2}, x_{3}, \cdots, x_{50}$ are filled, there is at least one column in the table where at least one number in that column is equal to the corresponding $x$. We call such a number a good number. Thus, $k$ is winning if and only if there exists a $k \times 50$ number table such that there is at least one good number in the table.

Clearly, $k=50$ is good. In fact, a $50 \times 50$ number table has 50 rows and 50 columns, with each row being a permutation of 1, 2, 3, ..., 50. The table contains 50 ones, and we can place these 50 ones in 50 columns, one in each column. This way, regardless of how $x_{1}, x_{2}, x_{3}, \cdots, x_{50}$ are filled, the 1 in the bottom ticket must be in the same column as one of the 1s in the table. Furthermore, $k=49$ is also good. We can construct a similar table as follows:
\begin{tabular}{|c|c|c|c|c|c|}
\hline 1 & 2 & 3 & 4 & *. & 49 \\
\hline 49 & 1 & 2 & 3 & $\cdot$ & 48 \\
\hline 48 & 49 & 1 & 2 & * & 47 \\
\hline … & ... & ... & ... & *. & ... \\
\hline 2 & 3 & 4 & 5 & ... & 1 \\
\hline
\end{tabular}

Since the numbers $1, 2, \cdots, 49$ in the bottom ticket cannot all be filled in the last position, at least one number must be filled in the first 49 positions, and it must match one of the corresponding numbers in the rows. Thus, we can let each number $i$ occupy each column, so that any number from 1 to 50 filled in the bottom ticket will match one of the numbers in the column. Therefore, we can construct the following number table:
\begin{tabular}{ccccccc}
1, & 2, & 3, & 4, & $\cdots$, & $a-1$, & $a$ \\
2, & 3, & 4, & 5, & $\cdots$, & $a$, & 1 \\
3, & 4, & 5, & 6, & $\cdots$, & 1, & 2 \\
$\cdots$ & $\cdots$ & $\cdots$ & $\cdots$ & $\cdots$, & $\cdots$, & $\cdots$ \\
$a$, & 1, & 2, & 3, & $\cdots$, & $a-2$, & $a-1$
\end{tabular}

And the numbers $1, 2, \cdots, a$ in the bottom ticket must have at least one filled in the first $a$ columns. To ensure that the numbers $1, 2, \cdots, a$ cannot all be filled in the last $50-a$ columns, we need $50-a < 25$. Therefore, $a \geqslant 26$. Thus, $k=26$ is winning.

Next, we prove that $k=25$ is not winning. In fact, consider a $25 \times 50$ number table. We can prove that we can fill a bottom ticket $P: x_{1}, x_{2}, x_{3}, \cdots, x_{50}$ such that there is no good number in the table. First, fill the number 1 in $P$. Since there are 25 ones in the table and 50 columns, there must be a column without a 1, and we can fill 1 in that column. We can fill $2, 3, \cdots, a-1$ in the same way until $a$ cannot be filled. At this point, $a \geqslant 26$. Otherwise, if $a < 26$, $P$ would have at most $a-1 \leqslant 24$ numbers filled, occupying 24 cells in $P$, but the table has only 25 $a$s, occupying 25 cells, totaling $24 + 25$ cells, so $P$ would have at least one cell left to fill $a$, which is a contradiction. Additionally, since the table has only 25 $a$s, occupying 25 columns, $P$ would have at least 25 cells left to fill $a$. But $a$ cannot be filled, meaning these 25 cells are occupied by 25 numbers from $1, 2, 3, \cdots, a-1$ (denoted as $x_{1}, x_{2}, x_{3}, \cdots, x_{25}$). Consider any empty cell $D$ in $P$. Since $D$'s column has only 25 numbers, $D$ can have at most 25 numbers that cannot be filled. Thus, at least one of $a, x_{1}, x_{2}, x_{3}, \cdots, x_{25}$ can be filled in $D$, but $a$ cannot be filled in $D$, so one of $x_{1}, x_{2}, x_{3}, \cdots, x_{25}$ must be filled in $D$. Then, fill $x_{i}$ in $D$ and fill $a$ in the original position of $x_{i}$, and the bottom ticket will have one more number filled. Continue this process until all empty cells in $P$ are filled, ensuring that no number in $P$ matches any number in its column.

In summary, the minimum value of $k$ is 26.