Three. (50 points) Given a set $A$ consisting of $3 n$ different positive integers. If the elements in set $A$ satisfy:
(1) The sum of any $n$ elements is no greater than $N$;
(2) The sum of any $n+1$ elements is greater than $N$.
Find the minimum value of $N$ (expressed in terms of $n$).
Let $A=\left\{a_{1}, a_{2}, \cdots, a_{3 n}\right\}$, where $a_{1}N .
$$

From equations (1) and (2), the above two inequalities can be written as
$$
\sum_{i=1}^{n}\left(a_{n+1}+n-1+i\right) \leqslant N, \sum_{j=1}^{n+1}\left(a_{n+1}+1-j\right)>N \text {. }
$$

Then $n a_{n+1}+\frac{n(3 n-1)}{2} \leqslant N$,
$$
\begin{array}{l}
(n+1) a_{n+1}-\frac{n(n+1)}{2}>N . \\
\text { Hence } \frac{2 N+n(n+1)}{2(n+1)}N .
\end{array}
$$

Therefore, the minimum value of $N$ is $\frac{4 n^{3}+3 n^{2}+n}{2}$.