Reverse and Subtract

Let the Number be $A =$ $a_0*10^ {2m} + a_1*10^{2m-1} + ... + a_k*10^{2m-k} + ... + a_{2m-1}*10 + a_{2m}$ with $n$ number of digits, where $n=2m+1$ , $m$ and $a_i$ belongs to positive integer set & $i = 0, 1, ... 2m$ belongs to non negative natural numbers .

Let the Reverse be $B =$ $a_{2m}*10^ {2m} + a_{2m-1}*10^{2m-1} + ... + a_k*10^{2m-k} + ... + a_{1}*10 + a_{0}$

Subtracting, we get, $A - B =$ $a_0* [10^ {2m} - 1] + a_1* [ 10^{2m-1} - 10 ^1 ] + ... + a_k*[ 10^{2m-k} - 10^{2m-k} ]$ where $k = m$

And We can prove that $10^{2m-r} - 10 ^r = \frac{ 10 ^ {2m} }{ 10 ^ {r} } - 10^{r} = \frac{ 10 ^ {2m} - 10 ^ {2r} }{10 ^ {r} } = \frac{ 10^{2r} (10 ^ {2(m-r)} - 1 ) }{10 ^ {r} }$

with $r < m => 2r < 2m$ & $(m-r) >=1$ since $r$ is a member of non-negative integer set .

Thus we get, $10^{2m-r} - 10 ^r = 10^{2r-r} (10 ^ {2(m-r)} - 1 )$ the general maximum of which is $99$ for all such $n$ . Thus, the answer is: $\boxed{99}$ .