II. (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 generated 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) $2 n-3$, (2) Obviously, any non-constant arithmetic sequence satisfies the requirement. Conversely, this is because: Let the $n$-element set $\left\{a_{1}, a_{2}, \cdots, a_{n}\right\}$ generate only $2 n-3$ distinct numbers, and $a_{1}<a_{2}<\cdots<a_{n}$. From $a_{1}+a_{n-1}<a_{1}+a_{n}<a_{2}+a_{n}$ and $a_{1}+a_{n-1}<a_{2}+a_{n-1}<a_{2}+a_{n}$, 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 $a_{2}+a_{n}=a_{3}+a_{n-1}$, and so on, it is easy to see that $a_{n}-a_{n-1}=a_{n-1}-a_{n-2}=\cdots=a_{2}-a_{1}$.