Let $A = \{1, 2, 3, \ldots, 20\}$.

$R_1 = \{(a, b) : b \text{ is divisible by } a\}$

$R_2 = \{(a, b) : a \text{ is an integral multiple of } b\}$

We want to find the number of elements in $R_1 - R_2$, which means we want to find the number of pairs $(a, b)$ such that $(a, b) \in R_1$ and $(a, b) \notin R_2$.

$(a, b) \in R_1$ means $b$ is divisible by $a$, so $b = ka$ for some integer $k$.

$(a, b) \in R_2$ means $a$ is an integral multiple of $b$, so $a = mb$ for some integer $m$.

$(a, b) \notin R_2$ means $a$ is not an integral multiple of $b$.

We want to find pairs $(a, b)$ such that $b = ka$ and $a \neq mb$ for any integer $m$.

If $b = ka$ and $a = mb$, then $b = k(mb) = kmb$, which implies $km = 1$. Since $k$ and $m$ are integers, we must have $k = m = 1$ or $k = m = -1$. Since $a, b \in A$, $a$ and $b$ are positive, so $k$ and $m$ must be positive. Thus, $k = m = 1$, which means $a = b$.

So, we want to find pairs $(a, b)$ such that $b = ka$ and $a \neq b$. This means $k \neq 1$.

We need to find the number of pairs $(a, b)$ such that $b = ka$ for some integer $k > 1$ and $a, b \in A$.

For $a = 1$, $b$ can be $2, 3, \ldots, 20$. There are 19 such values.

For $a = 2$, $b$ can be $4, 6, \ldots, 20$. There are 9 such values.

For $a = 3$, $b$ can be $6, 9, \ldots, 18$. There are 5 such values.

For $a = 4$, $b$ can be $8, 12, 16, 20$. There are 4 such values.

For $a = 5$, $b$ can be $10, 15, 20$. There are 3 such values.

For $a = 6$, $b$ can be $12, 18$. There are 2 such values.

For $a = 7$, $b$ can be $14$. There is 1 such value.

For $a = 8$, $b$ can be $16$. There is 1 such value.

For $a = 9$, $b$ can be $18$. There is 1 such value.

For $a = 10$, $b$ can be $20$. There is 1 such value.

For $a > 10$, $ka > 20$ for $k > 1$.

So the total number of pairs is $19 + 9 + 5 + 4 + 3 + 2 + 1 + 1 + 1 + 1 = 46$.