A positive integer gets reduced by 9 times when one of its digits is deleted....

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.

Write the first number as $10^{n+1}a+10^nb+c$ where $b$ is the digit that will be deleted, $c$ has $n$ digits, and $a$ can have multiple digits. We are told that $10^{n+1}a+10^nb+c=9(10^na+c)$ with $10^{n-1} \le c \lt 10^n$. This gives $8c=10^n(a+b)$, which shows $a+b \le 7$. The fact that deleting a digit does not spoil the divisibility by $9$ shows that $b=0$ as $b=9$ is prohibited. If we take $a=1,b=0$ we find the number to be $10125$ with as many trailing zeros as desired. Similarly we find the solutions $2025,30375,405,50626,6025,70875$, all of which can be multiplied by $10^k$. You delete the second digit $0$ to do the first division by $9$ and the first digit for the second division by $9$.