2. Given an integer $n \geqslant 3$. Find the smallest positive integer $k$, such that there exists a $k$-element set $A$ and $n$ pairwise distinct real numbers $x_{1}, x_{2}, \cdots, x_{n}$, satisfying $x_{1}+x_{2}, x_{2}+x_{3}, \cdots$, $x_{n-1}+x_{n}, x_{n}+x_{1}$ all belong to $A$. (Xiong Bin provided)
2. Let $x_{1}+x_{2}=m_{1}, x_{2}+x_{3}=m_{2}, \cdots \cdots$
$x_{n-1}+x_{n}=m_{n-1}, x_{n}+x_{1}=m_{n}$.
First, $m_{1} \neq m_{2}$, otherwise, $x_{1}=x_{3}$, which is a contradiction.
Similarly, $m_{i} \neq m_{i+1}\left(i=1,2, \cdots, n, m_{n+1}=m_{1}\right)$.
Thus, $k \geqslant 2$.
If $k=2$, let $A=\{a, b\}(a \neq b)$, such that
$$
\begin{array}{l}
\left\{\begin{array}{l}
x_{1}+x_{2}=a, \\
x_{2}+x_{3}=b, \\
\cdots \ldots . \\
x_{n-1}+x_{n}=b, \\
x_{n}+x_{1}=a
\end{array} \text { ( } n\right. \text { is odd), } \\
\text { or }\left\{\begin{array}{l}
x_{1}+x_{2}=a, \\
x_{2}+x_{3}=b, \\
\cdots \cdots \\
x_{n-1}+x_{n}=a, \\
x_{n}+x_{1}=b
\end{array} \quad(n \text { is even). }\right.
\end{array}
$$

For the system (1), $x_{n}=x_{2}$, which is a contradiction.
For the system (2),
$$
\begin{array}{l}
\frac{n}{2} a=\left(x_{1}+x_{2}\right)+\left(x_{3}+x_{4}\right)+\cdots+\left(x_{n-1}+x_{n}\right) \\
=\left(x_{2}+x_{3}\right)+\left(x_{4}+x_{5}\right)+\cdots+\left(x_{n}+x_{1}\right) \\
=\frac{n}{2} b,
\end{array}
$$

Thus, $a=b$, which is a contradiction.
For $k=3$, an example can be constructed as follows:
$$
\text { Let } x_{2 k-1}=k, x_{2 k}=n+1-k(k=1,2, \cdots) \text {. }
$$

Then, when $n$ is even,
$$
x_{\mathrm{i}}+x_{\mathrm{i}+1}=\left\{\begin{array}{ll}
n+1, & i \text { is odd; } \\
n+2, & i \text { is even and } i<n ; \\
\frac{n}{2}+2, & i=n\left(x_{n+1}=x_{1}\right) .
\end{array}\right.
$$

When $n$ is odd,
$$
x_{i}+x_{i+1}=\left\{\begin{array}{ll}
n+1, & i \text { is odd and } i<n ; \\
n+2, & i \text { is even; } \\
\frac{n-1}{2}+2, & i=n\left(x_{n+1}=x_{1}\right) .
\end{array}\right.
$$

In summary, the minimum value of $k$ is 3.