1. We need the number of 3-digit numbers divisible by $2$ and $3$, but not divisible by $4$ and $9$.

2. A number divisible by both $2$ and $3$ must be divisible by $\operatorname{lcm}(2,3)=6.$ So first count 3-digit multiples of $6$.

3. The 3-digit numbers run from $100$ to $999$.

The first 3-digit multiple of $6$ is $102$ and the last is $996$. Thus the count is $\frac{996-102}{6}+1=\frac{894}{6}+1=149+1=150.$

4. Now exclude those divisible by $4$ or $9$. Since we are already among multiples of $6$:

• divisible by $6$ and $4$ means divisible by $\operatorname{lcm}(6,4)=12$
• divisible by $6$ and $9$ means divisible by $\operatorname{lcm}(6,9)=18$

So from the 150 multiples of $6$, remove 3-digit multiples of $12$ or $18$.

5. Count 3-digit multiples of $12$: First is $108$, last is $996$. $\frac{996-108}{12}+1=\frac{888}{12}+1=74+1=75.$

6. Count 3-digit multiples of $18$: First is $108$, last is $990$. $\frac{990-108}{18}+1=\frac{882}{18}+1=49+1=50.$

7. Numbers counted in both are divisible by both $12$ and $18$, i.e. divisible by $\operatorname{lcm}(12,18)=36.$ Count 3-digit multiples of $36$: First is $108$, last is $972$. $\frac{972-108}{36}+1=\frac{864}{36}+1=24+1=25.$

8. By inclusion-exclusion, numbers divisible by $12$ or $18$ are $75+50-25=100.$

9. Therefore required count is $150-100=50.$

10. Equivalently, these are 3-digit numbers divisible by $6$ but by neither $12$ nor $18$.

So the final integer answer is $\boxed{50}.$

11. Comparison with stored answer:

• Derived answer: $50$
• Stored correct answer: $125$

These do not match. The stored answer appears incorrect.