4. Let the natural number $n$ have the following property: from $1,2, \cdots \cdots, n$ any 50 different numbers are taken, among these 50 numbers there must be two numbers whose difference is 7. The largest such $n$ is $\qquad$
4. 98

Solution: Since $50=7 \times 7+1$. By the Pigeonhole Principle, from $1,2, \cdots, 98$ any 50 different numbers chosen must have 8 numbers that, when divided by 7, yield the same remainder, let this remainder be $r$.

Also, since $98=14 \times 7$, there are exactly 14 numbers in $1,2, \cdots, 98$ that, when divided by 7, have a remainder of $r$. 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, from $1,2, \cdots, 99$ the following 50 numbers can be selected:
$$
\begin{array}{r}
1,2, \cdots, 7,15,16, \cdots, 21,29,30, \cdots, 35,43,44, \\
\cdots, 49,57,58, \cdots, 63,71,72, \cdots, 77,85,86, \cdots, 91,99
\end{array}
$$

Among these, no two numbers have a difference of 7, so 99 does not have the property mentioned in the problem. Therefore, the largest such $n$ is 98.