Solution:
We begin with a lemma.
Lemma. For all $k \geq 0$, $10^{k} \equiv 1 + 9k \pmod{81}$.
Proof. Binomial theorem:
$$10^{k} = (1 + 9)^{k} = 1^{k} + \binom{k}{1} 1^{k-1} 9 + \text{terms divisible by } 9^{2}.$$
We now write $N$ in terms of its digits as
$$\begin{aligned} N &= a_{0} + 10 a_{1} + \cdots + 10^{n} a_{n} \\ &\equiv (a_{0} + 9 \cdot 0 \cdot a_{0}) + (a_{1} + 9 \cdot 1 \cdot a_{1}) + \cdots + (a_{n} + 9 \cdot n \cdot a_{n}) \\ &\equiv a_{0} + a_{1} + \cdots + a_{n} + 9(0 a_{0} + 1 a_{1} + 2 a_{2} + \cdots + n a_{n}) \pmod{81} \end{aligned}$$
Correspondingly, the number formed by reversing the digits of $N$ is
$$a_{n} + 10 a_{n-1} + \cdots + 10^{n} a_{0} \equiv a_{0} + a_{1} + \cdots + a_{n} + 9(n a_{0} + (n-1) a_{1} + \cdots + 0 a_{n}) \pmod{81}.$$
If we add these two numbers, we get that $81$ divides
$$2(a_{0} + \cdots + a_{n}) + 9(n a_{0} + n a_{1} + \cdots + n a_{n}) = (9n + 2)(a_{0} + \cdots + a_{n}).$$
Since $9n + 2$ is not divisible by $3$, the conclusion follows.