Not all are 9's

The general formula for such $n$ digits number can be written as $99(10^{n-2}-1)$ Set $n =9$

The answer is $11\cdot (10^7-1)\cdot 9= 9899999901$ .

Additional facts $99(10^{n-2}-1) -11(10^{n-2}-1) = 88(10^{n-2}-1) = 8\cdot 11(10^{n-2}-1), \ \ n> 3$

Generalization: Let us try to figure out the smallest possible number such $\begin{array} {ccccc} \large & & b_0 & b_1 \cdots b_{n-1} & b_n \\ \large \times & & & &9 \\ \hline \large & & b_n & b_{n-1} \cdots b_1 & b_0 &\end{array}$ For a single digit number $9\cdot b_0 = b_0\implies b_0=0$ . On the similar manner we can observe that for two and three digit number $\begin{cases} 9(10b_0 + b_1) = 10b_1 +b_0 \Rightarrow 89b_0 = 90b_1\\ 9(100b_0+10b_1+b_2) = 100b_2+10b_1+b_0 \\ \Rightarrow 899b_0 + 80b_1 =91b_2 \\ \end{cases}$ For the above two cases we cannot find such $b_0 ,b_1 ,b_2\in\mathbb Z^+ < 10$ . Trying for the 4 digit number we can observe that $8999b_0 + 890b_1 -10b_2 - 991b_3 =0$ Clearly $b_0=1$ implying $b_3=9$ from above cryptogram set and $890b_1 -10b_2 =-80$ and hence we obtain $b_1=0 ,b_2=8$ and $1089$ our smallest number.

Note that $\begin{aligned} 11\cdot 99= 11(10^{4-2}-1) =1089\\ 11\cdot 999=11(10^{5-3}-1) = 10989 \\11\cdot 9999 = 11(10^{6-4}-1)= 109989 \end{aligned}$ from the above pattern $11(10^{n-2}-1)$ , we can easily determine the smallest nth digit number for $n\geq 4$ such property holds true. For 5 digit such number we set $n=5$ giving us $10989$ . Reversing the digit we further observe that $\begin{aligned} 9801 & =99\cdot 99 = 99(10^{4-2}-1) \\ 98901 & = 99\cdot 999= 99(10^{5-2}-1) \\ 989901 & =99\cdot 9999 = 99(10^{6-2}-1) \end{aligned}$ And we have then $99\cdot(10^{n-2}-1) ÷ 11(10^{n-2}-1) = 9.$ If we perform the subtraction we can observe that $99(10^n-1) -11(10^n-1) = 88(10^{n-2}-1) = 8\cdot 11(10^{n-2}-1)$ set $n= 2,3,4,\cdots$ then $\begin{aligned} 8\cdot 11(10^{4-2}-1) = 8712 \\ 8\cdot 11(10^{5-2}-1) = \boxed{87912}\\ 8\cdot 11(10^{6-2}-1) = 899712 \end{aligned}$ On reversing these numbers we can find the general formula $2\cdot 1089 = 2\cdot 11 (10^{n-2}-1)$ and hence $8\cdot 11(10^{n-2}-1) ÷ 2\cdot 11(10^{n-2}-1)=4.$

I had set up a problem upon these amazing number few weeks ago here Reflection .