The numerator simplifies to $(10^n-1)(10^n+1)$, and since $10^n+1$ is positive, we can factor it out from both the numerator and the denominator, which leaves us with $a=\frac{10^n-1}{3}$, whose digits sum to $567$. $10^n-1$ is a number of the form $999...999$, with $n$ total $9$s. As a result, $a=\frac{10^n-1}{3}$ is a number of the form $333...333$, with $n$ total $3$s. Therefore, we have $n=\frac{567}{3}=189$ $\blacksquare$