1 Positive integer $n$ satisfies the following property: when $n$ different odd numbers are chosen from $1,2, \cdots, 100$, there must be two whose sum is 102. Find the minimum value of $n$.
1. The minimum value of $n$ is 27. First, prove that $n \geqslant 27$. If $n \leqslant 26$, then let $A=\{1,3,5, \cdots, 2 n-1\}$, then $|A|=n$, but the sum of any two numbers in $A$: $(2 i-1)+(2 j-1)=2(i+j)-2 \leqslant 2(26+25)-2=100<102$ (where $0<i<j \leqslant n \leqslant 26$), which is a contradiction. Next, when $n=27$, let $A_{i}=\{2 i+1,101-2 i\}(i=1,2, \cdots, 24), A_{25}=\{1\}, A_{26}=\{51\}$. Place the $n \geqslant 27$ numbers into the above 26 sets, at least one set will contain two of these numbers, these two numbers can only be in some $A_{i}(1 \leqslant i \leqslant 24)$, and their sum is 102.