(50 points) Given a natural number $n \geqslant 5$. Try to find:
(1). In the $n$-element set $\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$, how many different numbers are produced by $a_{i}+a_{j}$ $(1<i<j \leqslant n)$ at least;
(2) Determine all $n$-element sets that achieve the above minimum value.
(1) Suppose $a_{1}<a_{2}<\cdots<a_{n}$, then it is clear that at least the following $2 n-3$ different numbers are generated:
$$
\begin{array}{l}
a_{1}+a_{2}<a_{1}+a_{3}<\cdots<a_{1}+a_{n}<a_{2}+a_{n} \\
<\cdots<a_{n-1}+a_{n} .
\end{array}
$$
(2) Clearly, a non-constant arithmetic sequence satisfies the requirement.

Conversely, this is also true. This is because:
Suppose the $n$-element set $\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ only generates $2 n-3$ different numbers, and $a_{1}<a_{2}<\cdots<a_{n}$.
From $a_{1}+a_{n-1}<a_{2}+a_{n-1}<a_{2}+a_{n}$, we get
$$
a_{1}+a_{n}=a_{2}+a_{n-1} \text {. }
$$

From $a_{1}+a_{n}=a_{2}+a_{n-1}<a_{3}+a_{n-1}<a_{3}+a_{n}$, we get
$$
a_{2}+a_{n}=a_{3}+a_{n-1} .
$$

Continuing in this manner, we have
$$
a_{i-1}+a_{n}=a_{i}+a_{n-1}, 1<i<n-1 .
$$

From $a_{1}+a_{n-2}<a_{2}+a_{n-2}<a_{2}+a_{n-1}=a_{1}+a_{n}$, we get $a_{1}+a_{n-1}=a_{2}+a_{n-2}$. Thus,
$$
a_{n}-a_{n-1}=a_{n-1}-a_{n-2}=\cdots=a_{2}-a_{1} .
$$