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 _______.

Let $A = \{1, 4, 7\}$ and $B = \{2, 3, 8\}$. The relation $R$ consists of pairs $((a_1, b_1), (a_2, b_2))$ from $A \times B$ such that $(a_1 + b_2)$ divides $(a_2 + b_1)$.

Let $x = a_1 + b_2$ and $y = a_2 + b_1$. Note that $(a_1, b_2)$ and $(a_2, b_1)$ are both independent elements of $A \times B$. So choosing $((a_1, b_1), (a_2, b_2))$ is equivalent to independently choosing two elements of $A \times B$ with sums $x$ and $y$ such that $x \mid y$.

All possible sums $a + b$ with $a \in A$ and $b \in B$:
$1 + 2 = 3$, $1 + 3 = 4$, $1 + 8 = 9$
$4 + 2 = 6$, $4 + 3 = 7$, $4 + 8 = 12$
$7 + 2 = 9$, $7 + 3 = 10$, $7 + 8 = 15$

Frequencies of each sum:
$f(3) = 1$, $f(4) = 1$, $f(6) = 1$, $f(7) = 1$, $f(9) = 2$, $f(10) = 1$, $f(12) = 1$, $f(15) = 1$

For each possible divisor $x$, count the number of valid pairs $(x, y)$ with $x \mid y$, each contributing $f(x) \cdot f(y)$:

$x = 3$: divides $3, 6, 9, 12, 15$
Ways $= 1 \cdot (1 + 1 + 2 + 1 + 1) = 6$

$x = 4$: divides $4, 12$
Ways $= 1 \cdot (1 + 1) = 2$

$x = 6$: divides $6, 12$
Ways $= 1 \cdot (1 + 1) = 2$

$x = 7$: divides $7$
Ways $= 1 \cdot 1 = 1$

$x = 9$: divides $9$
Ways $= 2 \cdot 2 = 4$

$x = 10$: divides $10$
Ways $= 1 \cdot 1 = 1$

$x = 12$: divides $12$
Ways $= 1 \cdot 1 = 1$

$x = 15$: divides $15$
Ways $= 1 \cdot 1 = 1$

Total number of elements in $R$:
$6 + 2 + 2 + 1 + 4 + 1 + 1 + 1 = 18$

Hence, the answer is $18$.