A piece of paper is flipped over, the numbers $0,1,8$ remain unchanged after a $180^{\circ}$ rotation, 6 becomes 9, 9 becomes 6, and other numbers have no meaning after a $180^{\circ}$ rotation. How many 7-digit numbers remain unchanged after a $180^{\circ}$ rotation? Among these, how many are divisible by 4? The total sum of these 7-digit numbers that remain unchanged after a $180^{\circ}$ rotation is $\qquad$
If $\overline{\mathrm{ABCDEFG}}$ remains unchanged after a $180^{\circ}$ rotation, then $D$ must be one of $0, 1, 8$. The pairs $C$ and $E$, $B$ and $F$, $A$ and $G$ can only be $0, 1, 8$ simultaneously, or one $6$ and one $9$. Therefore, we only need to determine the four digits $A, B, C, D$, so the total number of 7-digit numbers that remain unchanged after a $180^{\circ}$ rotation is $4 \times 5 \times 5 \times 3=300$.
Among these, the last two digits of the numbers that are divisible by 4 can only be $00, 08, 16, 60, 68, 80, 88, 96$. Among these, the numbers ending in $00, 80, 60$ will have a leading zero after a $180^{\circ}$ rotation, which is a contradiction, so they are excluded. Therefore, the numbers that are divisible by 4 are $5 \times 3 \times 5=75$.
The positions of $A, G$ each have $1, 8, 6, 9$ appearing $5 \times 5 \times 3=75$ times, the positions of $B, C, E, F$ each have all digits appearing $4 \times 5 \times 3=60$ times, and the position of $D$ has $0, 1, 8$ each appearing $4 \times 5 \times 5=100$ times. Therefore, the total sum of all 7-digit numbers that remain unchanged after a $180^{\circ}$ rotation is $24 \times 75 \times 1000001 + 24 \times 60 \times 110110 + 9 \times 100 \times 1000 = 1959460200$.