Step 1: Identify the structure of required numbers.
We need 4-digit numbers between 2000 and 5000 using digits $\{0,1,2,3,4\}$ without repetition. The thousands digit must be 2, 3, or 4 (it cannot be 0 since that gives a 3-digit number, and it cannot be 5 or above since we only have digits up to 4, and numbers must stay below 5000 with 4 in thousands giving exactly 4000s range which is still below 5000).
Step 2: Apply divisibility rule for 3.
A number is divisible by 3 if and only if the sum of its digits is divisible by 3. The total sum of all five available digits is $0+1+2+3+4=10$. Since we pick 4 out of 5 digits, exactly one digit is excluded. The sum of chosen digits equals $10 - (\text{excluded digit})$.
Step 3: Find which digit to exclude.
We need $10 - r \equiv 0 \pmod{3}$, which means $r \equiv 10 \equiv 1 \pmod{3}$. Among $\{0,1,2,3,4\}$, the digits congruent to 1 mod 3 are $r=1$ (since $1 \equiv 1$) and $r=4$ (since $4 \equiv 1$). So the excluded digit is either 1 or 4.
Step 4: Count valid numbers when digit 1 is excluded.
Remaining digits are $\{0,2,3,4\}$. The thousands digit can be 2, 3, or 4 (not 0), giving 3 choices. The remaining 3 positions are filled with the other 3 digits in any order: $3! = 6$ ways. Total for this case: $3 \times 6 = 18$.
Step 5: Count valid numbers when digit 4 is excluded.
Remaining digits are $\{0,1,2,3\}$. The thousands digit can be 2 or 3 only (not 0, and not 4 since 4 is excluded; also 1 in thousands gives 1000s which is below 2000). So thousands digit has 2 choices. The remaining 3 positions are filled with the other 3 digits: $3! = 6$ ways. Total for this case: $2 \times 6 = 12$.
Step 6: Add both cases for the final answer.
Total numbers = $18 + 12 = 30$. This matches option (2).
\[ \boxed{30} \]