Show that if $n = 9k \pm 3$, $k \in \mathbb{N}$, then the given equation has no natural solutions.
Let the decimal representations of $a$ and $b$ consist of $m$ and $l$ digits respectively. Then the initial equation is equivalent to the equation
$$\left(a + \frac{b}{10^l}\right) \left(b + \frac{a}{10^m}\right) = 9k \pm 3 \quad \text{or} \quad ab + \frac{ab}{10^{m+l}} + \frac{a^2}{10^m} + \frac{b^2}{10^l} = 9k \pm 3,$$
i. e.
$$10^{m+l}ab + ab + a^2 10^l + b^2 10^m = 10^{m+l} \cdot (9k \pm 3). \quad (*)$$
Since $10^t - 1 = \underbrace{9\dots9}_{t \text{ times}}$ for any positive integer $t$, i. e. $10^t = 9A + 1$ for some positive integer $A$, we can replace all powers of 10 in (*) by their presentations, then we obtain
$$(9A_1 + 1)ab + ab + (9A_2 + 1)a^2 + (9A_3 + 1)b^2 = (9A_4 + 1) \cdot (9k \pm 3)$$
or $(a+b)^2 = 9B \pm 3$ for some positive integer $B$. But this equality is impossible because the left hand side of it is the square number and the left hand side is a number which is divisible by 3 but is not divisible by 9.