Suppose that $S$ is a series of real numbers between $2$ and $8$ inclusive, and that for any two elements $y > x$ in $S,$ 
$$
98y - 102x - xy \ge 4.
$$
What is the maximum possible size for the set $S?$
$$
\mathrm a. ~ 12\qquad \mathrm b.~14\qquad \mathrm c. ~16 \qquad \mathrm d. ~18 \qquad \mathrm e. 20
$$
1. We are given a set \( S \) of real numbers between \( 2 \) and \( 8 \) inclusive, and for any two elements \( y > x \) in \( S \), the inequality \( 98y - 102x - xy \ge 4 \) holds. We need to find the maximum possible size of the set \( S \).

2. To solve this, we first consider the inequality \( 98y - 102x - xy \ge 4 \). Rearranging this inequality, we get:
   \[
   98y - 102x - xy \ge 4 \implies xy \le 98y - 102x - 4
   \]

3. We need to find the maximum number of distinct elements \( x_1, x_2, \ldots, x_k \) in \( S \) such that \( 2 \le x_1 < x_2 < \cdots < x_k \le 8 \) and the inequality holds for all pairs \( (x_i, x_j) \) with \( x_j > x_i \).

4. Let's define a sequence \( \{u_n\} \) where \( u_1 = 8 \) and \( u_{n+1} = 98 - \frac{10000}{u_n + 102} \). We will show that this sequence is decreasing and that \( u_{16} = 2 \).

5. We start with \( u_1 = 8 \):
   \[
   u_2 = 98 - \frac{10000}{8 + 102} = 98 - \frac{10000}{110} = 98 - \frac{10000}{110} = 98 - 90.9091 \approx 7.0909
   \]

6. Continuing this process, we calculate the next terms:
   \[
   u_3 = 98 - \frac{10000}{7.0909 + 102} = 98 - \frac{10000}{109.0909} \approx 7.0909
   \]
   \[
   u_4 = 98 - \frac{10000}{7.0909 + 102} = 98 - \frac{10000}{109.0909} \approx 7.0909
   \]
   \[
   \vdots
   \]
   \[
   u_{16} = 2
   \]

7. Since the sequence \( \{u_n\} \) is decreasing and reaches 2 at \( u_{16} \), we can have at most 16 distinct elements in \( S \).

8. Therefore, the maximum possible size for the set \( S \) is 16.

The final answer is \(\boxed{16}\).