Let $$A = \{1, 4, 7\}$$ and $$B = \{2, 3, 8\}$$.  Then the number of elements, in the relation $$R = \{((a_1, b_1), (a_2, b_2)) \in (A \times B) \times (A \times B) : a_1 + b_2 \text{ divides } a_2 + b_1\}$$. is :

The set of ordered pairs is $$A \times B = \{(1,2),(1,3),(1,8),(4,2),(4,3),(4,8),(7,2),(7,3),(7,8)\}$$, so there are $$9$$ elements in $$A \times B$$.

The relation $$R$$ is defined on $$(A \times B) \times (A \times B)$$ by
$$((a_1,b_1),(a_2,b_2)) \in R \iff a_1+b_2 \text{ divides } a_2+b_1.$$
Thus we have to count all ordered quadruples $$(a_1,b_1,a_2,b_2)$$ with this divisibility property.

Let $$d = a_1+b_2$$ (the divisor) and $$n = a_2+b_1$$ (the dividend).
For every choice of $$(a_1,b_2)$$ we get one value of $$d$$, and for every independent choice of $$(a_2,b_1)$$ we get one value of $$n$$. Hence

$$|R| = \sum_{d|n} (\text{number of ways to obtain } d)\times(\text{number of ways to obtain } n).$$

Compute all possible sums and their frequencies:

$$\begin{array}{c|ccc} a & b=2 & b=3 & b=8\\\hline 1 & 3 & 4 & 9\\ 4 & 6 & 7 & 12\\ 7 & 9 & 10 & 15 \end{array}$$

The multiset of sums and their counts is
$$\{3(1),\,4(1),\,6(1),\,7(1),\,9(2),\,10(1),\,12(1),\,15(1)\}.$$ (The number in parentheses is the frequency.)
Because the same table holds for both $$(a_1,b_2)$$ and $$(a_2,b_1)$$, the divisor and dividend distributions are identical.

List every possible divisor $$d$$, determine which dividends $$n$$ it divides, and accumulate the contributions:

$$\begin{array}{c|c|c} d & n\text{ that satisfy }d|n & \text{Contribution}=c_d\sum c_n\\\hline 3 & 3,6,9,12,15 & 1(1+1+2+1+1)=6\\ 4 & 4,12 & 1(1+1)=2\\ 6 & 6,12 & 1(1+1)=2\\ 7 & 7 & 1(1)=1\\ 9 & 9 & 2(2)=4\\ 10 & 10 & 1(1)=1\\ 12 & 12 & 1(1)=1\\ 15 & 15 & 1(1)=1 \end{array}$$

Adding all contributions:
$$|R| = 6+2+2+1+4+1+1+1 = 18.$$

Therefore, the required cardinality is 18.