4. Let the natural number $n$ have the following property: from $1,2,3, \cdots, n$ any 51 different numbers are taken, among these 51 numbers there must be two numbers whose sum is 101, what is the maximum value of such $n$?
4. Represent $n$ natural numbers as $n$ vertices, and connect an edge between the corresponding two vertices when the sum of two numbers is 101, to form a graph $G$. Thus, the problem is equivalent to finding the largest $n$ such that any selection of 51 vertices in $G$ must include two adjacent vertices.

When $51 \leqslant n \leqslant 100$, $G$ has exactly $n-50$ edges with no common vertices, and there are $50-(n-50)=100-n$ vertices of degree 0. Since the sum of the number of edges and the number of vertices of degree 0 is $(n-50)+(100-n)=50$, taking 51 vertices must include two adjacent vertices.

When $n>100$, we can choose the 50 even numbers $2, 4, 6, \cdots, 100$, plus 101, making a total of 51 numbers. Clearly, the sum of any two of these numbers is not 101.
Therefore, the $n$ that satisfies the condition is 100.