Example 1. (Wenzhou Junior High School Math Competition Question, 1987) Let the natural number $n$ have the following property: from 1, 2, ..., $n$, any 51 different numbers chosen will definitely have two numbers whose sum is 101. The largest such $n$ is $\qquad$
Consider $\{1,2, \cdots, n\}$ as the vertex set. When the sum of two numbers is 101, connect the corresponding two vertices to form a graph $G$. Clearly, the original problem is equivalent to finding the largest $n$ such that any selection of 51 vertices in $G$ must include two adjacent vertices.
i) 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 zero. Since the sum of the number of edges and the number of zero-degree vertices is $(n-50)+(100-n)=50$, taking 51 vertices must include two adjacent vertices.
ii) When $n>100$, $G$ has exactly 50 edges with no common vertices, and $n-100$ vertices of degree zero. Since $n>100$, we have $50+(n-100) \geqslant 51$. That is, in the set of 51 vertices formed by selecting one vertex from each of the 50 edges and adding any one of the zero-degree vertices, no two vertices are adjacent.
Combining i) and ii), the value of $n$ that meets the condition is 100.