4. A lottery ticket has 50 spaces arranged in sequence, and each participant should fill in the numbers 1 to 50 without repetition on their ticket, with the order determined by themselves. The host also fills in these 50 numbers on a ticket as the answer key. If a participant's number in any space matches the number in the corresponding space on the answer key, they win the lottery. How many tickets should a participant fill out at least to ensure they will definitely win (i.e., no matter how the answer key is filled, they will have at least one winning ticket)?
4. 26 tickets. To ensure a win, for example, 26 tickets can be filled out as follows:
$$
\begin{array}{l}
1,2,3, \cdots, 25,26,27, \cdots, 50 \\
2,3,4, \cdots, 26,1,27, \cdots, 50 \\
3,4,5, \cdots, 1,2,27, \cdots, 50 \\
\cdots \cdots
\end{array}
$$
$$
\begin{array}{l}
25,26,1, \cdots, 23,24,27, \cdots, 50 \\
26,1,2, \cdots, 24,25,27, \cdots, 50
\end{array}
$$
At least one number will appear in the first 26 positions (since there are only 24 positions in the back). Therefore, this number must match one of the 26 tickets filled out. To prove that filling out 25 tickets may not guarantee a win, place the 25 filled tickets and one blank ticket one by one. Let's see how to fill the blank ticket so that the numbers filled in do not match any of the 25 filled tickets.

First, it is clear that 1 can be filled in an appropriate position so that its position does not match the 1 on any of the 25 tickets. Assuming that 1, 2, ..., $a-1$ have already been filled, let's see how to fill $a$. If it can be filled and is in a different position from the $a$ on the 25 tickets, then do so. If it cannot be filled, it means that all suitable positions for $a$ are already occupied. Originally, there should be no fewer than 25 suitable positions for $a$; this is because each ticket can only create one "forbidden position." Suppose the suitable positions for $a$ on the blank ticket have already been filled with $x_{1}, x_{2}, \cdots, x_{28}$. Take any empty position; clearly, there are no fewer than 25 suitable numbers for that position, so at least one of $x_{1}, x_{2}, \cdots, x_{28}$ and $a$ can be filled in that position. Since $a$ is not suitable for that position, the position is suitable for filling some number $x_{1}$. Fill $x$ in that position, and then fill $a$ in the original position of $x_{1}$ ${ }^{1}$. Continue this process until the end, and you will get a blank ticket that ensures no ticket can win.