Solution: The key claim is the following.
Claim: The sum in the problem is equal to the number of solutions of $2 x+3 y \leq 300$ where $x, y$ are positive integers.

Proof. The sum in the problem is the same as counting the number of triples $(a, b, d)$ of positive integers such that $\operatorname{gcd}(a, b)=1$ and $d(2 a+3 b) \leq 300$. Now, given such $(a, b, d)$, we biject it to the pair $(x, y)$ described in the claim by $x=d a$ and $y=d b$. This transformation can be reversed by $d=\operatorname{gcd}(x, y)$, $a=x / d$, and $b=y / d$, implying that it is indeed a bijection, so the sum is indeed equal to the number of such $(x, y)$.

Hence, we wish to count the number of positive integer solutions to $2 x+3 y \leq 300$. One way to do this is via casework on $y$, which we know to be an integer less than 100:

- If $y$ is even, then $y=2 k$ for $1 \leq k \leq 49$. Fixing $k$, there are exactly $\frac{300-6 k}{2}=150-3 k$ values of $x$ which satisfy the inequality, hence the number of solutions in this case is
$$\sum_{k=1}^{49}(150-3 k)=\frac{150 \cdot 49}{2}=3675$$
- If $y$ is odd, then $y=2 k-1$ for $1 \leq k \leq 50$. Fixing $y$, there are exactly $\frac{302-6 k}{2}=151-3 k$ values of $x$ which satisfy the inequality, hence the number of solutions in this case is
$$\sum_{k=1}^{50}(151-3 k)=\frac{149 \cdot 50}{2}=3725$$
The final answer is $3675+3725=7400$.