Write $a,b,c$ for the sums of each of the three rows. We have $a+b+c=1+\cdots+9=45$. The fact that each row is a multiple of 9 means that $a,b,c\in\{9,18,27,36\}$. So $a,b,c$ are either $27,9,9$ or $18,18,9$, in some order. As $1+2+3+4=10$, the 4-digit term cannot be $9$, so it is $18$ or $27$. But $27$ is not the sum of four consecutive integers, so the third row is $3,4,5,6$ in some order.
We are left with $1,2,7,8,9$ for the other two rows, one adding to $18$ and the other $9$. We cannot reach $18$ with two digits at all or with three digits not using the $9$, so the first row is necessarily one of the following two possibilities: $9,8,1$ in some order, and the middle one $2,7$; or $2,7,9$ in some order and $1,8$. This last case is easily discarded since the only 4-digit product is $279\times18=5022$, which produces the wrong last row.
Looking at the first row, the last digit cannot be $1$ because that would imply a repeated digit. $2$ and $9$ do not fit, because they would put an $8$ in the last row. Actually, if the first row starts with $8$ or $9$, as $819\times27$ already has five digits, we learn that the first number starts with $1$ and that the second one is $27$.
We have $189\times27=5103$, not good, so it has to be $198\times 27=5346$. So the answer to the question is $8+7+6=21$.
Having established that the original number is divisible by 9, the deleted digit must be either 0 or 9 because only these two digits can leave the digit sum divisible by 9. Call the deleted digit $d$ and assume it is in the $10^k$ place:
$$aaa\dots adb\dots bbb=A\cdot10^{k+1}+d\cdot10^k+B$$
$$aaa\dots ab\dots bbb=A\cdot10^k+B$$
$$A\cdot10^{k+1}+d\cdot10^k+B=9(A\cdot10^k+B)$$
$$10A\cdot10^k+d\cdot10^k+B=9A\cdot10^k+9B$$
$$A\cdot10^k+d\cdot10^k=8B$$
$$8B=(A+d)\cdot10^k$$
$$B=(A+d)\cdot\frac{10^k}8$$
Yet by our construction above we must have $B<10^k$, so
$$(A+d)\cdot\frac{10^k}8<10^k$$
$$A+d<8$$
If $d=9$ then $A$ would be forced to be negative, which is impossible. Therefore $d=0$, $A<8$ and all numbers satisfying the conditions in the first part of the question are of the form
$$N=A\cdot10^{k+1}+A\cdot\frac{10^k}8,\ 0<A<8,\ k\ge3-\log_2\gcd(A,8)$$
The restriction on $k$ ensures that $A\cdot\frac{10^k}8$ is an integer. $A$ cannot be zero because $N$ would then start with a zero.
The numbers $N$ fall into seven classes depending on what $A$ is:
$$A=1: N=10125\cdot10^{k-3}$$
$$A=2: N=2025\cdot10^{k-2}$$
$$A=3: N=30375\cdot10^{k-3}$$
$$A=4: N=405\cdot10^{k-1}$$
$$A=5: N=50625\cdot10^{k-3}$$
$$A=6: N=6075\cdot10^{k-2}$$
$$A=7: N=70875\cdot10^{k-3}$$
Regardless of what $k$ is, division by 9 will not touch the trailing zeros, so we can ignore them. Dividing $N$ by 9 removes the zero that is second from left, producing the following prefixes, and dividing by 9 again can be accomplished by deleting the leftmost digit:
$$\require{cancel}A=1:\cancel1125\ldots\to125\dots$$
$$A=2:\cancel225\ldots\to25\dots$$
$$A=3:\cancel3375\ldots\to375\dots$$
$$A=4:\cancel45\ldots\to5\dots$$
$$A=5:\cancel5625\ldots\to625\dots$$
$$A=6:\cancel675\ldots\to75\dots$$
$$A=7:\cancel7875\ldots\to875\dots$$
Hence the proof asked for by the question, that $\frac N{81}$ can be reached by deleting a single digit from $\frac N9$, has been shown.
Best Answer
Each number in base ten can be written as $$d_0\cdot 10^0+d_1\cdot 10^1+\cdots+d_n10^n$$ where each $d_i$ is between $0$ and $9$ inclusive. (We would write the number in base ten as the concatenation $d_nd_{n-1}\cdots d_0$).
But note that $10\equiv 1 \pmod 9$, and thus every power of ten is $1$ modulo $9$ (that is, the remainder of a power of ten when divided by $9$ must be $1$).
So if we write our number modulo $9$, we have $$d_0\cdot 10^0+d_1\cdot 10^1+\cdots+d_n10^n=d_0+d_1+\cdots+d_n.$$ Thus, if our number is divisible by $9$, it must be congruent to $0$ modulo $9$, and so the sum of the digits must be divisible by $9$.
Edit: If you have never seen congruences and modular arithmetic, this article should be helpful. It is simpler than it sounds. We write $a\equiv b\pmod n$ if $b$ is the remainder of $a$ when divided by $n$.