Let the decimal representation of $a$ consist of $k$ digits. Set $b = 5 \cdot 10^m$, $m \in \mathbb{N}$. Then
$$\begin{align*} \overline{a,b} \cdot \overline{b,a} &= n \iff \left(a + \frac{b}{10^k}\right)\left(b + \frac{a}{10^m}\right) = n \iff \\ &\iff \frac{(2a + 1) \cdot 5 \cdot 10^m}{2} + \frac{a(2a + 1)}{2 \cdot 10^k} = n \iff \\ &\iff (2a + 1) \cdot 5^{m+1} \cdot 2^{m-1} + \frac{a(2a + 1)}{2 \cdot 10^k} = n. \end{align*}$$
We see that the last equality holds for some positive integer $n$ if and only if the number $a(2a + 1)$ is divisible by $2 \cdot 10^k = 2^{k+1} \cdot 5^k$. So if we find at least one such number $a$, then the problem statement will be proved, since for this $a$ and any $b = 5 \cdot 10^m$, $m \in \mathbb{N}$ the number $a, b \cdot b, a$ will be integer.
Consider $k=4$. Then the required relation has the form $a(2a+1) \mid 2^5 \cdot 5^4$.
It is easily followed from two relations $a \mid 2^5$ and $(2a+1) \mid 5^4$. So, to solve the problem it suffices to find at least one $a$ with 4 digits in its decimal representation and satisfying these two relations. Such $a$ does exist, for example, $a = 5312$. Show how we can find such $a$.
First we show that there exist infinitely many positive integers $a$ such that $a \mid 2^5$ and $(2a+1) \mid 5^4$. The first relation is equivalent to the equality $a = 32c$, where $c \in \mathbb{N}$, and then the second relation has the form $2 \cdot 32c + 1 \mid 5^4$ or $64c + 1 \mid 625$. We have
$$\begin{align*} 64c + 1 \mid 625 &\iff 64c + 1 - 625 \mid 625 &\iff 64c - 624 \mid 625 &\iff \\ 16(4c - 39) \mid 625 &\iff 4c - 39 \mid 625 &\iff 4c - 39 - 625 \mid 625 &\iff \\ 4c - 664 \mid 625 &\iff 4(c - 166) \mid 625 &\iff c - 166 \mid 625 &\iff \\ c = 625l + 166, l \in \mathbb{Z}. \end{align*}$$
In particular, for $l = 0$ we have $c = 166$, and so $a = 32c = 32 \cdot 166 = 5312$ has 4 digits in its decimal representation, as required.