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 edge of a small rectangle, if both endpoints are nodes and its interior does not contain any other nodes, it is called a "basic segment". Considering all possible divisions, find the maximum and minimum number of basic segments. ${ }^{[4]}$
（32nd China Mathematical Olympiad）
【Analysis】This problem can be seen as a plane region being divided into 2017 pieces, and we need to find the maximum and minimum values of the number of edges. Using Euler's theorem, the number of edges can be converted into calculating the number of points.

The number of points can be considered from two aspects: one is from the point of view, each point can be the vertex of 1, 2, or 4 small rectangles; the other is from the perspective of small rectangles, each small rectangle has four vertices. Therefore, by calculating the sum of all vertices of each small rectangle, an estimate of the maximum value can be obtained. On the other hand, from the perspective of each horizontal and vertical line segment dividing the square, the two endpoints of each horizontal and vertical line segment are points with a degree of 3. In summary, an estimate of the minimum number of points can be obtained.

Solution: Construct a graph $G$ with all nodes as vertices and basic line segments as edges. Let the number of vertices of graph $G$ be $v$, and the number of edges be $e$. Consider the external region of rectangle $R$ as one face (region). Then graph $G$ has a total of $f=2017$ faces.
By Euler's theorem, we have $v+f-e=2$.
Thus, $e=v+2015$.
Note that, the points with a degree of 2 in graph $G$ are exactly 4, and the rest of the points have a degree of 3 or 4. A point with a degree of 2 is a vertex of one rectangle, a point with a degree of 3 is a vertex of two rectangles, and a point with a degree of 4 is a vertex of four rectangles.

On one hand, the sum of all vertices of each small rectangle is $4 \times 2016$.

On the other hand, the sum of all vertices of all small rectangles is not less than $2(v-4)+4=2 v-4$.
Then $2 v-4 \leqslant 4 \times 2016 \Rightarrow v \leqslant 4034$.
Therefore, from equation (1), we get $e \leqslant 6049$.
When the rectangle $R$ is divided into $1 \times 2016$ small rectangles, the above equation holds with equality.
Thus, the maximum value sought 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, where these horizontal and vertical lines do not include the boundaries.

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

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

Since each horizontal and vertical line segment has two endpoints that are vertices of two rectangles, they are all points with a degree of 3, and these points are not repeated.
Thus, $x \geqslant 2 a+2 b$.
Combining with equation (2), we get
$$
\begin{array}{l}
x \geqslant 2(a+1)+2(b+1)-4 \\
\geqslant 4 \sqrt{(a+1)(b+1)}-4 \\
\geqslant 4 \sqrt{2016}-4>175.59 \\
\Rightarrow x \geqslant 176 .
\end{array}
$$

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

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