Consider $n = \overline{a_1a_2\dots a_{m-1}a_m}$ and $r(n) = \overline{a_m a_{m-1} \dots a_2 a_1}$, its reverse.

$$n + r(n) = \sum_{j=0}^{m} (a_j + a_{m-j}) \cdot 10^j = \sum_{j=0}^{m} a_j (10^j + 10^{m-j}).$$

Notice that $10^i + 10^{j+1} \equiv 10^j + 10^{i+1} \pmod{81}$, $\forall i, j \in \mathbb{N}$. It follows that there exists $r$ such that $10^j + 10^{m-j} \equiv r$, $\forall j = 0, m$. We obtain that $n + r(n) \equiv r \sum_{j=0}^{m} a_j$. As $(r, 81) = 1$, the conclusion follows immediately.