3. Rectangle $R$ is divided into 2016 small rectangles, with each small rectangle's sides parallel to the sides of rectangle $R$. The vertices of the small rectangles are called "nodes". A line segment on the side of a small rectangle, if both of its endpoints are nodes and its interior does not contain any other nodes, is called a "basic segment". Considering all possible divisions, find the maximum and minimum number of basic segments.

As shown in Figure 2, rectangle $R$ is divided into five small rectangles, with a total of 16 basic segments. Segments $A B, B C$ are basic segments, while segment $A C$ is not a basic segment.
(Qiu Zhenhua, problem contributor)
3. Construct a graph $G$ with all nodes as vertices and basic segments as edges. Let there be $N$ basic segments, which is the number of edges in graph $G$. It is easy to see that the degree of vertices in graph $G$ can only be $2$, $3$, or $4$.

Vertices with degree 2 are exactly the four vertices of rectangle $R$.

Let the number of vertices with degree $3$ and $4$ be $x$ and $y$, respectively. By the relationship between the number of edges and vertex degrees in a graph, we have
$$
N=\frac{1}{2}(4 \times 2 + 3x + 4y) = 4 + \frac{3}{2}x + 2y.
$$

Each small rectangle has four vertices, and considering repeated counting, there are $2016 \times 4$ vertices in total. Vertices with degree 2 are counted once, vertices with degree 3 are counted twice, and vertices with degree 4 are counted four times.
Thus, $2016 \times 4 = 4 + 2x + 4y$
$\Rightarrow x + 2y = 4030$.
Substituting equation (2) into equation (1) gives
$$
N = 4034 + \frac{1}{2}x.
$$

From equation (3), finding the maximum and minimum values of $N$ is equivalent to finding the maximum and minimum values of $x$, and by equation (2), it is also equivalent to finding the minimum and maximum values of $y$.

Consider dividing rectangle $R$ into 2016 small rectangles using 2015 vertical segments. In this case, there are no vertices with degree 4, so $y$ takes the minimum value $0$, and $x$ takes the maximum value 4030.
Therefore, $N_{\max} = 6049$.
Next, we find the minimum value of $N$.
Let all basic segments inside rectangle $R$ lie on $s$ horizontal lines and $t$ vertical lines. Then, rectangle $R$ can be divided into at most $(s+1)(t+1)$ small rectangles.
Thus, $(s+1)(t+1) \geqslant 2016$.
Suppose there is a basic segment $e$ on a vertical line $l$, extending $e$ to two nodes until it cannot be extended further (extending further would enter other small rectangles or exit rectangle $R$), these two nodes are vertices of degree 3 in graph $G$, shaped like "丁" and "⊥". Similarly, each horizontal line containing basic segments inside rectangle $R$ also corresponds to at least two vertices of degree 3, shaped like "—" and "†", and these vertices of degree 3 are distinct.
$$
\begin{array}{l}
\text{Thus, } x \geqslant 2s + 2t. \\
\text{By } s + t = (s+1) + (t+1) - 2 \\
\geqslant 2 \sqrt{(s+1)(t+1)} - 2 \\
\geqslant 2 \sqrt{2016} - 2 > 87 \\
\Rightarrow s + t \geqslant 88. \\
\text{Substituting into equation (4) gives } x \geqslant 176. \\
\text{Substituting into equation (3) gives } N \geqslant 4122.
\end{array}
$$

Consider dividing rectangle $R$ into $45^2 = 2025$ small rectangles using 44 horizontal lines and 44 vertical lines, then merging any ten consecutive small rectangles in the first row into one small rectangle. In this case, rectangle $R$ is divided into 2016 small rectangles, and the number of vertices with degree 3 is exactly 176, so $N$ takes the minimum value $N_{\min} = 4122$.