(2) Divide the sides of the equilateral $\triangle A B C$ with side length 1 into $n$ equal parts, and draw lines parallel to the other two sides through each division point, dividing the triangle into smaller triangles. The vertices of these small triangles are called nodes. A real number is placed at each node. It is known that
( I ) The numbers placed at points $A, B, C$ are $a, b, c$;
(II) In each rhombus formed by two smallest triangles sharing a common side, the sums of the numbers placed at the two pairs of opposite vertices are equal.
Find:
(1) The shortest distance $r$ between the node with the maximum number and the node with the minimum number;
(2) The total sum $S$ of the numbers on all nodes.
Without loss of generality, let $a \leqslant b \leqslant c$. We refer to property (II) as the "diamond property". As shown in Figure 1, consider three adjacent nodes $D, E, F$ on the same line, with numbers $d, e, f$ placed on them; then take node $G$ adjacent to $D, E$, and node $H$ adjacent to $E, F$, with numbers $g, h$ placed on them. According to the "diamond property", we have $g+e=d+h, g+f=e+h$, so $f-e=e-d$. This indicates that the numbers on $D, E, F$ form an arithmetic sequence.
By the "transitivity" of arithmetic sequences, it is easy to see that all nodes on the same line form an arithmetic sequence; by the "boundary property" of arithmetic sequences, it is easy to see that if the common difference is non-zero, the maximum (minimum) value of the arithmetic sequence appears at the ends. Therefore:
(1) If $a<b<c$, it is easy to see that the maximum (minimum) number on each line is at the ends, so point $C$ places the maximum number $c$, point $A$ places the minimum number $a, r=1$. If $a=b<c$, similarly, point $C$ places the maximum number $c$, any node on line $AB$ places the minimum number $a$, the one closest to $C$ is $P$, when $n$ is even, $P$ is the midpoint of $AB$; when $n$ is odd, $P$ is the $\frac{n-1}{2}$ or $\frac{n+1}{2}$th $n$-th division point of $AB$, for the former $r=\frac{\sqrt{3}}{2}$, for the latter $r=\sqrt{\left(\frac{\sqrt{3}}{2}\right)^{2}+\left(\frac{1}{2 n}\right)^{2}}=\frac{1}{2 n} \sqrt{3 n^{2}+1}$. If $a<b=c$, the result is the same; if $a=b=c$, then all nodes place the same number, which is both the maximum and minimum number, so $r=0$. In summary,
$$
r=\left\{\begin{array}{l}
1, a, b, c \text { are all different } \\
\frac{\sqrt{3}}{2}, a, b, c \text { have two equal, and } n \text { is even } \\
\frac{1}{2 n} \sqrt{3 n^{2}+1}, a, b, c \text { have two equal, and } n \text { is odd } \\
0, a=b=c
\end{array}\right.
$$
(2) Rotate $\triangle ABC$ around its center by $120^{\circ}, 240^{\circ}, 360^{\circ}$, and superimpose the three triangles obtained from the three rotations into one $\triangle A^{\prime} B^{\prime} C^{\prime}$. The numbers placed on the nodes of $\triangle A^{\prime} B^{\prime} C^{\prime}$ are the sums of the numbers superimposed at the same node. Since arithmetic sequences have "additivity" (the sum of two arithmetic sequences is still an arithmetic sequence), it is easy to see that the numbers on all nodes on each line of $\triangle A^{\prime} B^{\prime} C^{\prime}$ still form an arithmetic sequence, but at this time the numbers placed on the three vertices $A^{\prime}, B^{\prime}, C^{\prime}$ are all $a+b+c$, so all nodes place the same number $a+b+c$. The total sum of the numbers on all nodes of $\triangle A^{\prime} B^{\prime} C^{\prime}$ is $3S=N(a+b+c)$, where $N$ is the total number of nodes, i.e.

So
$$
\begin{array}{c}
N=1+2+\cdots+(n+1) \\
S=\frac{1}{6}(n+1)(n+2)(a+b+c)
\end{array}
$$