Let $$A = \{n \in N : H.C.F.(n, 45) = 1\}$$ and let $$B = \{2k : k \in \{1, 2, \ldots, 100\}\}$$. Then the sum of all the elements of $$A \cap B$$ is ______

$$A = \{n \in \mathbb{N} : \text{HCF}(n, 45) = 1\}$$ and $$B = \{2k : k \in \{1, 2, \ldots, 100\}\} = \{2, 4, 6, \ldots, 200\}$$.

We need the sum of all elements of $$A \cap B$$.

Since $$45 = 3^2 \times 5$$, requiring $$\text{HCF}(n, 45) = 1$$ means that $$n$$ is not divisible by 3 or 5. Therefore $$A \cap B$$ consists of the even numbers from 1 to 200 that are not divisible by 3 or 5.

The sum of even numbers in $$\{1, \ldots, 200\}$$ is $$ S = 2 + 4 + \cdots + 200 = 2 \cdot \frac{100 \cdot 101}{2} = 10100. $$

Now subtract the sum of even multiples of 3 (i.e., multiples of 6):

$$ 6, 12, \ldots, 198:\quad \text{count} = 33,\quad \sum = 6 \cdot \frac{33 \cdot 34}{2} = 6 \cdot 561 = 3366. $$

Next subtract the sum of even multiples of 5 (i.e., multiples of 10):

$$ 10, 20, \ldots, 200:\quad \text{count} = 20,\quad \sum = 10 \cdot \frac{20 \cdot 21}{2} = 2100. $$

From this add back the sum of even multiples of 15 (i.e., multiples of 30):

$$ 30, 60, \ldots, 180:\quad \text{count} = 6,\quad \sum = 30 \cdot \frac{6 \cdot 7}{2} = 630. $$

Substituting these into the inclusion-exclusion formula gives $$ \text{Sum} = 10100 - 3366 - 2100 + 630 = 5264. $$

The answer is $$\boxed{5264}$$.