To determine the number of elements in $R_1 - R_2$, let's first articulate the meaning of both relations on set $A = \{1, 2, 3, \ldots, 20\}$:

$R_1$ includes pairs $(a, b)$ where $b$ is divisible by $a$. This includes pairs like $(1,1), (1,2), \ldots, (1,20)$ for $1$; similar series for $2$ up to $(2,20)$ (excluding odd numbers); for $3$ up to $(3,18)$; and so on, reflecting the divisibility condition.

$\begin{aligned} & R_1:\left\{\begin{array}{l}(1,1),(1,2) \ldots,(1,20), \\ (2,2),(2,4) \ldots,(2,20), \\ (3,3),(3,6) \ldots,(3,18), \\ (4,4),(4,8) \ldots,(4,20), \\ (5,5)(5,10) \ldots,(5,20), \\ (6,6),(6,12),(6,18),(7,7),(7,14), \\ (8,8),(8,16),(9,9),(9,18)(10,10), \\ (10,20),(11,11),(12,12) \ldots,(20,20)\end{array}\right\} \\\\ & n\left(R_1\right)=66\end{aligned}$

$R_2$ consists of pairs $(a, b)$ where $a$ is an integral multiple of $b$. Essentially, this relationship is the reverse of $R_1$. However, for the essence of $R_1 - R_2$, the key overlap comes with pairs where $a = b$, since those are the pairs that clearly reflect both conditions identically (i.e., a number is always divisible by itself, and it is always a multiple of itself). There are 20 such pairs corresponding to each integer in $A$ from 1 to 20, resulting in the common elements between $R_1$ and $R_2$ (the intersection $R_1 \cap R_2$) being 20 pairs.

$\begin{aligned} & \mathrm{R}_1 \cap \mathrm{R}_2=\{(1,1),(2,2), \ldots(20,20)\} \\\\ & \mathrm{n}\left(\mathrm{R}_1 \cap \mathrm{R}_2\right)=20 \end{aligned}$

The difference $R_1 - R_2$ seeks elements present in $R_1$ but not in $R_2$. Given that $R_1$ and $R_2$ share 20 elements that are identical, to find $R_1 - R_2$, we subtract these 20 common elements from the total in $R_1$, resulting in $66 - 20 = 46$ pairs.

$\begin{aligned} & \mathrm{n}\left(\mathrm{R}_1-\mathrm{R}_2\right)=\mathrm{n}\left(\mathrm{R}_1\right)-\mathrm{n}\left(\mathrm{R}_1 \cap \mathrm{R}_2\right) \\\\ & =\mathrm{n}\left(\mathrm{R}_1\right)-20 \\\\ & =66-20 \\\\ & \mathrm{R}_1-\mathrm{R}_2=46 \text { Pair }\end{aligned}$