1 to 9

Each digit appears in each column $8!=40320$ times.

So each column must add up to $40320(1+2+3+4+5+6+7+8+9)=181400$

Therefore, the sum of all complete numbers is

$sum=1814400(1+10+100+1000+10000+100000+1000000+10000000+100000000)=$ $\large \color{#D61F06}\boxed{201599999798400}$

There exists a complete number that adds up to another complete number such that they equal to $1111111110$

Consider the complete number $123456789$ . Now, observe that $987654321$ "matches" with each digit in $123456789$ . Taking their sum lends us $1111111110$ . Another example would be $967485231$ which pairs up with $143625879$ and adds up to $1111111110$ . So, there will always be 2 complete numbers such that they sums up to $1111111110$ .

There are $9!$ complete numbers in total, hence there are exactly $\frac{9!}{2}$ pairs, each having a value of $1111111110$

Therefore, the sum of all complete number is:

$1111111110 \times \frac{9!}{2} = \large \boxed{201599999798400}$