2.34 Let the natural number $n$ have the following property: from $1,2, \cdots, n$, any 50 different numbers chosen will have at least two numbers whose difference is 7. Find the maximum value of such an $n$.
(China Junior High School Mathematics League, 1987)
[Solution] Since $50=7 \cdot 7+1$, by the pigeonhole principle, from the 98 numbers $1,2,3, \cdots, 97,98$, any 50 numbers chosen must have 8 numbers that have the same remainder when divided by 7. Let the remainder be $r$, where $r \in\{0,1,2, \cdots, 6\}$.

Since $98=14 \cdot 7$, there are exactly 14 numbers in $1,2,3, \cdots, 98$ that have a remainder of $r$ when divided by 7. Among these 14 numbers, any 8 different numbers must have two numbers whose difference is 7, meaning 98 has the property mentioned in the problem.
On the other hand, in $1,2,3, \cdots, 99$, we can choose the following 50 numbers.
$$
\begin{array}{l}
1,2,3,4,5,6,7, \\
15,16,17,18,19,20,21, \\
29,30,31,32,33,34,35, \\
43,44,45,46,47,48,49, \\
57,58,59,60,61,62,63, \\
71,72,73,74,75,76,77, \\
85,86,87,88,89,90,91, \\
99 .
\end{array}
$$

Among these, the difference between any two numbers is not 7, so the 99 numbers $1,2, \cdots, 99$ do not have the property mentioned in the problem.
Therefore, the maximum value of $n$ that satisfies the requirement is 98.