1. We need the sum of all distinct four-digit numbers formed using the digits $2,1,2,3$.

2. Since the digit $2$ is repeated twice, the number of distinct 4-digit numbers is $\frac{4!}{2!}=12.$

3. To find the total sum, use the symmetry idea:

• In all distinct permutations, each position (thousands, hundreds, tens, units) is occupied equally often by each digit, accounting for repetition.

Let us count how many times each digit appears in any fixed position.

• Total permutations = $12$.

• For digit $1$: fixing $1$ in a chosen position, the remaining digits are $2,2,3$, which can be arranged in $\frac{3!}{2!}=3$ ways. So digit $1$ appears $3$ times in each position.

• For digit $3$: similarly, fixing $3$ in a chosen position, the remaining digits are $1,2,2$, which can be arranged in $\frac{3!}{2!}=3$ ways. So digit $3$ appears $3$ times in each position.

• For digit $2$: there are two identical $2$'s. Fix one position with $2$; the remaining digits are $1,2,3$, which can be arranged in $3!=6$ ways. Hence digit $2$ appears $6$ times in each position.

Check: total appearances per position $3+6+3=12,$ which is correct.

4. So the sum of digits appearing in any one position is $1\cdot 3 + 2\cdot 6 + 3\cdot 3 = 3+12+9=24.$

5. Therefore total sum of all numbers is $24\times (1000+100+10+1).$ Now, $1000+100+10+1=1111.$ So, $\text{Sum}=24\times 1111=26664.$

6. Hence the required sum is $\boxed{26664}.$