Example 6 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". For a line segment on the side of a small rectangle, if both endpoints are nodes and its interior does not contain any other nodes, then this line segment is called a "basic segment". Considering all possible divisions, find the maximum and minimum number of basic segments. ${ }^{[4]}$
Consider a graph $G$ with all nodes as vertices and basic segments as edges. Let the number of vertices in graph $G$ be $v$, and the number of edges be $e$. Treat the external region of rectangle $R$ as one face (region). Then, the total number of faces in graph $G$ is $f=2017$.
By Euler's formula, we have $v+f-e=2$.
Thus, $e=v+2015$.
Note that, the points with degree 2 in graph $G$ are exactly 4, the rest of the points have a degree of 3 or 4. Points with degree 2 are the vertices of one rectangle, points with degree 3 are the vertices of two rectangles, and points with degree 4 are the vertices of four rectangles.

On one hand, the total number of vertices of all small rectangles is $4 \times 2016$.

On the other hand, the total number of vertices of all small rectangles is no less than $2(v-4)+4=2 v-4$.
Thus, $2 v-4 \leqslant 4 \times 2016 \Rightarrow v \leqslant 4034$.
Therefore, from equation (1), we get $e \leqslant 6049$.
When the rectangle $R$ is divided into $1 \times 2016$ small rectangles, the above equality holds.
Thus, the maximum value of $e$ is $e_{\text {max }}=6049$.
Next, consider the minimum value of $e$.
Suppose the rectangle $R$ is divided using $a$ horizontal lines and $b$ vertical lines, excluding the boundaries.

Since $a$ horizontal lines and $b$ vertical lines can divide the area into at most $(a+1)(b+1)$ regions, we have
$$
(a+1)(b+1) \geqslant 2016 \text {. }
$$

Let the number of points with degree 3 be $x$, and the number of points with degree 4 be $y$. Then,
$$
v=4+x+y, 4+2 x+4 y=4 \times 2016 .
$$

Each horizontal and vertical line has its endpoints as the vertices of two rectangles, thus they are all points with degree 3, and these points are distinct.
Hence, $x \geqslant 2 a+2 b$.
Combining with equation (2), we get
$$
\begin{array}{l}
x \geqslant 2(a+1)+2(b+1)-4 \\
\geqslant 4 \sqrt{(a+1)(b+1)}-4 \\
\geqslant 4 \sqrt{2016}-4>175.59 \\
\Rightarrow x \geqslant 176 .
\end{array}
$$

Thus, from equations (3) and (4), we get
$$
v=2019+\frac{1}{2} x \geqslant 2017 \text {. }
$$

From equation (1), we immediately get $e \geqslant 4122$, where the equality holds in the $42 \times 48$ division.
Thus, $e_{\min }=4122$.