Call the number $\overline{a_1a_2... a_m}$ ($a_1 \ne 0,a_m \ne 0$) the reverse of the number $\overline{a_m...a_2a_1}$. Prove that the sum between a number $n$ and its reverse is a multiple of $81$ if and only if the sum of the digits of $n$ is a multiple of $81$.
Claim: Let $t$ ,$m$ and $n$ be positive integers with $t \ge m \ge n$. Then $10^t+1 \equiv 10^{t-m}+10^m \equiv 10^n+10^{t-n} mod(81)$
Proof: We already know that $9|10^x-1$ for all x non-negative integer. So $81|(10^{m-n}-1)(10^n-1)$ thus the claim is true. (I guess it is trivial)
So the rest is very clear. Let $x \equiv 10^n+10^{m-n} (mod 81)$ for every n. Then since $\overline{a_0a_1... a_m}+\overline{a_ma_{m-1}... a_0}= \sum_{i=0}^{m} ((10^{m-i}+10^i)a_i) \equiv x(a_0+a_1+...+a_m) (mod 81) $ and $(x,81)=1$ the conclusion follows. $\blacksquare$