China Western Mathematical Olympiad

Let $m_1=x_1+x_2$, $m_2=x_2+x_3$, $\dots$, $m_{n-1}=x_{n-1}+x_n$, $m_n=x_n+x_1$. First, note that $m_1\ne m_2$, otherwise $x_1=x_3$, which contradicts the fact that $x_i$ are distinct. Similarly, $m_i\ne m_{i+1}$, for $i=1,2,\dots ,n$, where $m_{n+1}=m_1$, as usual. It follows that $k\ge 2$.

For $k=2$, let $A=\{a,b\}$, where $a\ne b$. It follows that $$(1)\{\begin{array}{l}{x}_1+x_2=a,\\ x_2+x_3=b,\\ ⋮\\ x_{n-1}+x_n=a,\\ x_n+x_1=b,\end{array}\text{(if n is even)}$$ or $$(2)\{\begin{array}{l}{x}_1+x_2=a,\\ x_2+x_3=b,\\ ⋮\\ x_{n-1}+x_n=b,\\ x_n+x_1=a.\end{array}\text{(if n is odd)}$$ For (2), we have $x_n=x_2$, which is possible. For (1), it follows that $$\begin{array}{rl}\frac{n}{2}a& =(x_1+x_2)+(x_3+x_4)+\cdots +(x_{n-1}+x_n)\\ & =(x_2+x_3)+(x_4+x_5)+\cdots +(x_n+x_1)\\ & =\frac{n}{2}b,\end{array}$$ and hence $a=b$, which is impossible again. It follows that $k\ge 3$.

For $k=3$, one can construct a valid example as follows: Define $x_{2k-1}=k$ ($k\ge 1$) and $x_{2k}=n+1-k$ ($k\ge 1$). When $n$ is even, $$x_i+x_{i+1}=\{\begin{array}{ll}n+1,& \text{if }i\text{ is odd,}\\ n+2,& \text{if }i\text{ is even and }i<n,\\ \frac{n}{2}+2,& \text{if }i=n,\text{ where }{x}_{n+1}=x_n.\end{array}$$ When $n$ is odd, $$x_i+x_{i+1}=\{\begin{array}{ll}n+1,& \text{if }i\text{ is odd and }i<n,\\ n+2,& \text{if }i\text{ is even,}\\ \frac{n-1}{2}+2,& \text{if }i=n,\text{ where }{x}_{n+1}=x_n.\end{array}$$ Therefore, the smallest positive integer $k$ is $3$.