The sum of the two numbers is $36$ and their HCF is $4$. How many pairs of these numbers are possible?

Let the two numbers be $a$ and $b$.
Since their Highest Common Factor (HCF) is $4$, we can express the numbers as:
$a=4x$ and $b=4y$
where $x$ and $y$ are co-prime positive integers, meaning $\text{HCF}(x,y)=1$.

According to the question, the sum of the numbers is $36$:

$$a+b=36$$

$$4x+4y=36$$

$$4(x+y)=36$$

$$x+y=9$$

Now we need to find the pairs of positive integers $(x,y)$ such that $x+y=9$ and $\text{HCF}(x,y)=1$:

1. $x=1,y=8⟹\text{HCF}(1,8)=1$ (Valid)
2. $x=2,y=7⟹\text{HCF}(2,7)=1$ (Valid)
3. $x=3,y=6⟹\text{HCF}(3,6)=3\ne 1$ (Invalid)
4. $x=4,y=5⟹\text{HCF}(4,5)=1$ (Valid)

The combinations $(5,4),(7,2),(8,1)$ give the same sets of numbers.
Thus, there are exactly $3$ unique possible pairs of numbers:

* $(4\times 1,4\times 8)=(4,32)$
* $(4\times 2,4\times 7)=(8,28)$
* $(4\times 4,4\times 5)=(16,20)$

So, the number of possible pairs is $3$.