3.42 The lottery has 50 spaces arranged in sequence, and each participant should fill in the 50 natural numbers from 1 to 50 on their lottery ticket without repetition, with the order determined by themselves. The host also fills in these 50 natural numbers on a lottery ticket as the base. In a lottery ticket, if any of the 50 numbers filled in matches the number in the same position on the base ticket, it is considered a win. How many lottery tickets should a participant fill out at minimum to ensure that they have at least one winning ticket?
[Solution] Fill 26 lottery tickets as follows:
$$
\begin{array}{l}
1,2,3, \cdots, 25,26,27, \cdots, 50, \\
2,3,4, \cdots, 26,1,27, \cdots, 50, \\
3,4,5, \cdots, 1,2,27, \cdots, 50, \\
\cdots \quad \cdots \quad \cdots \\
25,26,1, \cdots, 23,24,27, \cdots, 50, \\
26,1,2, \cdots, 24,25,27, \cdots, 50,
\end{array}
$$

That is, the first 26 natural numbers are rotated among the 26 lottery tickets, while the last 24 numbers are always filled in the last 24 positions unchanged. On the underlying lottery ticket, one of the numbers $1,2, \cdots, 26$ will always appear in one of the first 26 positions, thus it must match the position of one of the 26 lottery tickets, meaning that one of the 26 lottery tickets will definitely win.

Next, we will prove that if only 25 lottery tickets are filled, it cannot be guaranteed that one of them will win, i.e., there is always a possible underlying ticket that will result in none of the 25 lottery tickets winning.

Assume that 25 lottery tickets have already been filled. We take 1 blank lottery ticket and first fill in “1” in an appropriate position so that it does not match the position of “1” on any of the 25 filled lottery tickets. Then, we sequentially fill in $2,3,4, \cdots$ such that each number does not match the position of the same number on the 25 filled lottery tickets, until it is no longer possible to fill in any more numbers. Clearly, there are only a finite number of different ways to fill in the numbers, so there must be a filling method $M$ that allows the maximum number of natural numbers to be filled in while ensuring that no number matches the position of the same number on the 25 filled lottery tickets. We can prove that all 50 natural numbers must be filled into this lottery ticket.

If not, suppose that in the filling method $M$, the natural numbers $1,2, \cdots, a-1$ have all been filled in, but $a$ cannot be filled in, where $25 < a \leq 50$. This means that at least 25 positions suitable for filling $a$ have already been occupied. Let the numbers in these 25 positions be $x_{1}, x_{2}, \cdots, x_{25}$. Take any empty position, then among the 26 numbers $x_{1}, x_{2}, \cdots, x_{25}$ and $a$, at least one can be filled in, say $x_{i}$. Thus, we can fill $x_{i}$ into this empty position and fill $a$ into the position occupied by $x_{i}$. This means that this filling method can fill in one more number than $M$, which is impossible. Therefore, the filling method $M$ must fill all 50 numbers into the lottery ticket, and when this lottery ticket is used as the underlying ticket, none of the 25 filled lottery tickets will win.