Alter Digit

The digit that is deleted must be a $0$ or a $9,$ because the digit sum must be divisible by 9 both before and after the deletion.

If the digit is a $9,$ the situation is that $A = 10^{k+1} x + 9 \cdot 10^k + y,$ where $y < 10^k,$ and $B = 10^k x + y$ is the post-deletion integer, which is $1/9$ of $A.$ So we get $\begin{aligned} 9(10^k x + y) &= 10^{k+1} x + 9 \cdot 10^k + y \\ 8y &= 9 \cdot 10^k + 10^k x \end{aligned}$ which is manifestly impossible, since the right side is less than $8 \cdot 10^k.$

If the digit is a $0,$ we similarly get $A = 10^{k+1}x + y,$ $B = 10^k x + y,$ and $\begin{aligned} 9(10^k x + y) &= 10^{k+1} x + y \\ 8y &= 10^k x \end{aligned}$ Since $y < 10^k,$ we must have $x<8.$ So there are exactly $7$ values of $x$ to consider. In each case, note that $y$ will be divisible by $5,$ so we'll need $y$ to be odd so that $A$ is not divisible by $10.$

For $x=1$ we get $y = 10^k/8$ is an odd integer, so $k=3, y = 125, A = 10125, B = 1125.$

For $x=2$ we get $y=10^k/4$ is an odd integer, so $k=2, y = 25, A = 2025, B = 225.$

For $x=3$ we get $y = 3 \cdot 10^k/8$ is an odd integer, so $k=3, y = 375, A = 30375, B = 3375.$

For $x=4$ we get $y= 10^k/2$ is an odd integer, so $k=1, y = 5, A = 405, B = 45.$

For $x=5$ we get $y=5 \cdot 10^k/8$ is an odd integer, so $k=3, y = 625, A = 50625, B = 5625.$

For $x=6$ we get $y=3 \cdot 10^k/4$ is an odd integer, so $k=2, y = 75, A = 6075, B = 675.$

For $x=7$ we get $y=7 \cdot 10^k/8$ is an odd integer, so $k=3, y = 875, A = 70875, B = 7875.$

The sum of all the $A$ 's is $\fbox{170505}.$