1. Interpret the sets

   We have $A = \{n \in \mathbb{N} : \gcd(n,45)=1\}$ and $B = \{2k : k \in \{1,2,\dots,100\}\} = \{2,4,6,\dots,200\}.$

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

2. Condition for belonging to $A$

   Since $45 = 3^2 \cdot 5,$ the condition $\gcd(n,45)=1$ means that $n$ must be divisible by neither $3$ nor $5$.

3. Elements of $B$

   Every element of $B$ is of the form $2k$, where $1 \le k \le 100$.

   So we need all even numbers from $2$ to $200$ that are not divisible by $3$ and not divisible by $5$.

4. Sum of all elements of $B$

   $B = 2,4,6,\dots,200$ is an AP with 100 terms.

   Hence $S_B = 2(1+2+\cdots+100)=2\cdot \frac{100\cdot 101}{2}=10100.$

5. Subtract elements divisible by $3$

   In $B$, a number $2k$ is divisible by $3$ iff $k$ is divisible by $3$.

   So such terms are $6,12,18,\dots,198,$ i.e. $6\times 1, 6\times 2, \dots, 6\times 33$.

   Their sum is $6(1+2+\cdots+33)=6\cdot \frac{33\cdot 34}{2}=3366.$

6. Subtract elements divisible by $5$

   In $B$, a number $2k$ is divisible by $5$ iff $k$ is divisible by $5$.

   So such terms are $10,20,30,\dots,200,$ i.e. $10\times 1,10\times 2,\dots,10\times 20$.

   Their sum is $10(1+2+\cdots+20)=10\cdot \frac{20\cdot 21}{2}=2100.$

7. Add back elements divisible by both $3$ and $5$

   These are the elements divisible by $\operatorname{lcm}(3,5)=15.$

   In $B$, they are $30,60,90,120,150,180,$ i.e. $30\times 1,30\times 2,\dots,30\times 6$.

   Their sum is $30(1+2+\cdots+6)=30\cdot \frac{6\cdot 7}{2}=630.$

8. Apply inclusion-exclusion

   Required sum $= 10100 - 3366 - 2100 + 630.$

   Compute: $10100 - 3366 = 6734,$ $6734 - 2100 = 4634,$ $4634 + 630 = 5264.$

9. Final answer

   $\boxed{5264}$

10. Comparison with stored answer

Stored correct answer = $5264$.

This matches our derived answer.