Consider a rectangle $R$ partitioned into $2016$ smaller rectangles such that the sides of each smaller rectangle are parallel to one of the sides of the original rectangle. We aim to find the maximum and minimum possible number of basic segments over all possible partitions of $R$.

Let $s_i$ be the number of vertices which are intersections of $i$ segments in the partition. Let $N$ be the number of basic segments in the partition. Let $a$ and $b$ be the number of vertical and horizontal interior lines, respectively, which contain a segment in the partition.

Clearly, $s_2 = 4$, representing the four corners of $R$. Each vertex which is an intersection of $i$ segments belongs to $i$ basic segments. Also, every basic segment belongs to two vertices. Hence,
$$2N = 2s_2 + 3s_3 + 4s_4 = 8 + 3s_3 + 4s_4 \quad (1).$$

Each vertex which is an intersection of $i$ segments belongs to $1, 2, 4$ rectangles, where $i = 2, 3, 4$ respectively. Also, every rectangle belongs to four vertices. Hence,
$$4 \cdot 2016 = s_2 + 2s_3 + 4s_4 = 4 + 2s_3 + 4s_4,$$
which simplifies to
$$4030 = s_3 + 2s_4 \quad (2).$$

Now, subtracting twice equation (2) from equation (1), we get:
$$2N - 8060 = 8 + 3s_3 + 4s_4 - 2s_3 - 4s_4 = 8 + s_3,$$
which simplifies to
$$N = 4034 + \frac{s_3}{2} \quad (3).$$

From equation (2), we obtain that $s_3 \leq 4030$. Hence,
$$N = 4034 + \frac{s_3}{2} \leq 4034 + \frac{4030}{2} = 6049.$$
The maximum of $6049$ is achieved, for example, when $R$ is partitioned into $1 \times 2016$ rectangles. Hence, the maximum number of basic segments is $6049$.

If we draw an extension of every interior segment until it meets the boundary of $R$, we get a new partition into $(a+1) \times (b+1)$ rectangles, and we clearly increase the total number of rectangles. Hence,
$$(a+1)(b+1) \geq 2016 \quad (4).$$

Also, if we extend every interior segment as far as possible along borders between rectangles, we finish at two vertices which are intersections of $3$ edges. All these endpoints are clearly distinct. Hence,
$$s_3 \geq 2(a+b) \quad (5).$$

Using equations (3), (4), (5), and applying the AM-GM inequality, we get:
$$2016 \leq (a+1)(b+1) \leq \left( \frac{a+b}{2} + 1 \right)^2 \leq \left( \frac{s_3}{4} + 1 \right)^2,$$
which implies
$$s_3 + 4 \geq \lceil \sqrt{32256} \rceil = 180,$$
thus,
$$s_3 \geq 176.$$
Therefore,
$$N = 4034 + \frac{s_3}{2} \geq 4034 + \frac{176}{2} = 4122.$$
The minimum of $4122$ is achieved, for example, when $R$ is partitioned into $42 \times 48$ rectangles. Hence, the minimum number of basic segments is $4122$.

The answer is: $\boxed{4122 \text{ (minimum)}, 6049 \text{ (maximum)}}$.