2. 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 generated by $a_{i}+a_{j}(1 \leqslant i<j \leqslant n)$ at least?
(2) Determine all $n$-element sets that achieve the above minimum value.
2. (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: $a_{1}+a_{2}<a_{1}+a_{3}<\cdots<a_{1}+a_{n}<a_{2}+a_{n}<\cdots<a_{n-1}+a_{n}$.
(2) Clearly, an arithmetic sequence that is not constant satisfies the requirement.
Conversely, 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 $a_{1}+a_{n}=a_{2}+a_{n-1}$.
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 $a_{1}+a_{n-1}=a_{2}+a_{n-2}$, thus, $a_{n}-a_{n-1}=a_{n-1}-a_{n-2}=\cdots=a_{2}-a_{1}$.