The sum of all the 4-digit distinct numbers that can be formed with the digits 1, 2, 2 and 3 is:

We need to find the sum of all 4-digit distinct numbers formed using the digits 1, 2, 2, 3. Since two digits are identical (the two 2's), we must be careful about counting distinct arrangements.

The total number of 4-digit arrangements of {1, 2, 2, 3} is $$\frac{4!}{2!} = 12$$ distinct numbers.

To find their sum, we calculate the contribution of each digit to each place value (thousands, hundreds, tens, units). For any fixed position, we count how many times each digit appears there across all 12 arrangements.

For digit 1: We fix 1 in a position, and arrange {2, 2, 3} in the remaining 3 positions, giving $$\frac{3!}{2!} = 3$$ arrangements. So digit 1 appears 3 times in each position.

For digit 3: Similarly, fix 3 in a position and arrange {1, 2, 2}, giving $$\frac{3!}{2!} = 3$$ arrangements. So digit 3 appears 3 times in each position.

For digit 2: Since there are two 2's, fix one 2 in a position and arrange the remaining {1, 2, 3} in 3 positions, giving $$3! = 6$$ arrangements. But since both 2's are identical, this already counts correctly. So digit 2 appears 6 times in each position.

We can verify: $$3 + 6 + 3 = 12$$ total appearances per position, which matches our 12 distinct numbers.

The digit sum contribution per position is $$3 \times 1 + 6 \times 2 + 3 \times 3 = 3 + 12 + 9 = 24$$.

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

The answer is Option A: $$26664$$.