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 $m_1 \neq m_2$, otherwise $x_1 = x_3$, which contradicts the fact that $x_i$ are distinct. Similarly, $m_i \neq m_{i+1}$, 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 $a \neq b$. It follows that
$$(1) \begin{cases} x_1 + x_2 = a, \\ x_2 + x_3 = b, \\ \vdots \\ x_{n-1} + x_n = a, \\ x_n + x_1 = b, \end{cases} \quad \text{(if $n$ is even)}$$
or
$$(2) \begin{cases} x_1 + x_2 = a, \\ x_2 + x_3 = b, \\ \vdots \\ x_{n-1} + x_n = b, \\ x_n + x_1 = a. \end{cases} \quad \text{(if $n$ is odd)}$$
For (2), we have $x_n = x_2$, which is possible. For (1), it follows that
$$\begin{aligned} \frac{n}{2}a &= (x_1 + x_2) + (x_3 + x_4) + \dots + (x_{n-1} + x_n) \\ &= (x_2 + x_3) + (x_4 + x_5) + \dots + (x_n + x_1) \\ &= \frac{n}{2}b, \end{aligned}$$
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,
$$x_i + x_{i+1} = \begin{cases} n+1, & \text{if } i \text{ is odd,} \\ n+2, & \text{if } i \text{ is even and } i < n, \\ \frac{n}{2} + 2, & \text{if } i = n, \text{ where } x_{n+1} = x_n. \end{cases}$$
When $n$ is odd,
$$x_i + x_{i+1} = \begin{cases} n+1, & \text{if } i \text{ is odd and } i < n, \\ n+2, & \text{if } i \text{ is even,} \\ \frac{n-1}{2} + 2, & \text{if } i = n, \text{ where } x_{n+1} = x_n. \end{cases}$$
Therefore, the smallest positive integer $k$ is $3$.