1. Let the natural number $n$ have the following property: from $1,2, \cdots, n$, any 51 different numbers are chosen. Among these 51 numbers, there must be two numbers whose sum is 101. What is the largest such $n$?
1. Consider $\{1,2, \cdots, n\}$ as the set of vertices. 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. It is easy to see that from any vertex, there is at most one edge.

When $51 \leqslant n \leqslant 100$, $G$ has exactly $n-50$ edges, and there are $n-2(n-50)=100-n$ vertices with degree 0. Since $n \geqslant 51$, we have $100-n \leqslant 49$. Therefore, when selecting 51 vertices, there must be two adjacent vertices.

When $n>100$, $G$ has 50 edges and $n \cdots 100$ vertices with degree 0. Since $n>100$, we have $n-100>0$. By selecting one vertex from each of the 50 edges and adding any one of the zero-degree vertices, we form a set of 51 vertices that are pairwise non-adjacent.
In summary, the value of $n$ that meets the condition is 100.