1. Form all distinct 4-digit numbers using digits $1,2,2,3$

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

2. Use place value symmetry

In all distinct arrangements of the multiset $\{1,2,2,3\}$, each position (thousands, hundreds, tens, units) gets each digit equally often, accounting for repetition properly.

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

• Total distinct numbers $=12$
• Total number of positions filled across all numbers $=12\times 4=48$

Now:

• Digit $1$ appears once in each number, so across all numbers it appears $12$ times total.
• Digit $3$ appears once in each number, so across all numbers it appears $12$ times total.
• Digit $2$ appears twice in each number, so across all numbers it appears $24$ times total.

By symmetry, these appearances are equally distributed among the 4 places:

• $1$ appears $12/4=3$ times in each place.
• $3$ appears $12/4=3$ times in each place.
• $2$ appears $24/4=6$ times in each place.

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

3. Compute total sum

Each place contributes:

• thousands place: $24\times 1000$
• hundreds place: $24\times 100$
• tens place: $24\times 10$
• units place: $24\times 1$

Hence total sum is $24(1000+100+10+1)=24\times 1111=26664.$

4. Check options

$26664$ matches Option A.

Therefore, the required sum is $\boxed{26664}.$