1. We need to form 4-digit numbers between $1000$ and $3000$ using digits $1,2,3,4,5,6$ without repetition, and the number must be divisible by $4$.

2. Since the number is between $1000$ and $3000$, the thousands digit can only be: $1 \text{ or } 2$

3. For divisibility by $4$, the last two digits must form a number divisible by $4$.

   Using digits $1,2,3,4,5,6$ without repetition, let us list all possible 2-digit endings divisible by $4$:

   $12,16,24,32,36,52,56,64$

   (We exclude $44$ since repetition is not allowed.)

4. Now count valid 4-digit numbers casewise.

Case 1: Thousands digit = $1$

The ending (last two digits) cannot contain $1$.

From the list $12,16,24,32,36,52,56,64$ remove those containing $1$: $24,32,36,52,56,64$ So there are $6$ possible endings.

For each such ending, the hundreds digit can be chosen from the remaining digits.

After fixing the thousands digit and the last two digits, $3$ digits remain, so the hundreds digit has: $3 \text{ choices}$

Hence total numbers in this case: $6 \times 3 = 18$

Case 2: Thousands digit = $2$

The ending cannot contain $2$.

From the list $12,16,24,32,36,52,56,64$ remove those containing $2$: $16,36,52,56,64$ So there are $5$ possible endings.

Again, after fixing thousands digit and last two digits, $3$ digits remain for the hundreds digit.

Hence total numbers in this case: $5 \times 3 = 15$

5. Therefore total valid numbers: $18+15=33$

6. So the required number of integers is: $\boxed{33}$

7. Comparison with stored answer:

• Derived answer = $33$
• Stored correct answer = $30$

They do not match.

8. Quick verification by direct listing of valid endings:

• For thousands digit $1$: endings $24,32,36,52,56,64$ $(6)$, each gives $3$ choices for the hundreds digit $\Rightarrow 18$.
• For thousands digit $2$: endings $16,36,52,56,64$ $(5)$, each gives $3$ choices for the hundreds digit $\Rightarrow 15$.

Thus total is indeed: $18+15=33$

So the stored answer $30$ appears to be incorrect.