$2 \cdot 42$ Let the natural number $n$ have the following property: when 50 different numbers are chosen from the natural numbers $1, 2, \cdots, n$, there must be two numbers whose difference is 7. Find the maximum value of $n$.
[Solution] Starting from 1, take 7 consecutive natural numbers, then skip 7 natural numbers and start again from 15 to take 7 consecutive natural numbers, and continue this process until 98, obtaining a total of 49 natural numbers. Adding 99, we get 50 natural numbers:
$$
14 i+j, i=0,1, \cdots, 6, j=1,2, \cdots, 7 \text { and } i=7, j=1 .
$$

It is easy to see that the difference between any two of these 50 numbers is not equal to 7. Therefore, the maximum value of $n$ is less than 99.
When $n=98$, divide the 98 numbers into 7 sets:
$$
\begin{array}{l}
\{1,8,15, \cdots, 85,92\}, \\
\{2,9,16, \cdots, 86,93\}, \\
\{3,10,17, \cdots, 87,94\}, \\
\{4,11,18, \cdots, 88,95\}, \\
\{5,12,19, \cdots, 89,96\}, \\
\{6,13,20, \cdots, 90,97\}, \\
\{7,14,21, \cdots, 91,98\},
\end{array}
$$

Each set contains exactly 14 numbers, and the difference between any two adjacent numbers is 7. Any 50 chosen numbers are distributed among these 7 sets. By the pigeonhole principle, there must be at least one set containing at least 8 of the chosen numbers. Among these 8 numbers, there must be two adjacent numbers, whose difference is 7.
In summary, the maximum value of $n$ is 98.