Three, (25 points) Try to determine, for any $n$ positive integers, the smallest positive integer $n$ such that at least 2 of these numbers have a sum or difference that is divisible by 21.


Three, let $A_{i}$ represent all numbers that leave a remainder of $i$ when divided by 21; $i=0,1$, $\cdots, 20$; let $B_{j}=A_{j} \cup A_{21-j}, j=1,2, \cdots, 10, B_{11}=A_{0}$.

By taking 12 numbers, there will be at least 2 numbers belonging to the same category among the 11 categories $B_{1}, B_{2}$, $\cdots, B_{11}$. Therefore, their sum or difference must be a multiple of 21.
Thus, the smallest positive integer $n$ that satisfies the condition is 12.