Now there's an $n$ ?

Note first that the sum of a series of logs (of the same base) is equal to the log of the product of those terms. Now we can write this product as

$\dfrac{a(a + 2)}{(a - 2)(a + 4)}*\dfrac{(a + 1)(a + 3)}{(a - 1)(a + 5)}*\dfrac{(a + 2)(a + 4)}{a(a + 6)}*\dfrac{(a + 3)(a + 5)}{(a + 1)(a + 7)}*.....$

This is a telescoping product that leaves us with the equation

$\log_{n}\left(\dfrac{(a + 2)(a + 3)}{(a - 2)(a - 1)}\right) = 1 \Longrightarrow (a + 2)(a + 3) = n(a - 2)(a - 1)$

$\Longrightarrow (n - 1)a^{2} - (3n + 5)a + 2(n - 3) = 0$

$\Longrightarrow a = \dfrac{(3n + 5) \pm \sqrt{(3n + 5)^{2} - 8(n - 1)(n - 3)}}{2(n - 1)}$

$\Longrightarrow a = \dfrac{(3n + 5) \pm \sqrt{n^{2} + 62n + 1}}{2(n - 1)}.$

Now in order for $a$ to be rational with $n$ being integral we will require that the polynomial under the square root sign be a perfect square, i.e.,

$n^{2} + 62n + 1 = m^{2} \Longrightarrow (n + 31)^{2} - 960 = m^{2}$

$\Longrightarrow [(n + 31) + m][(n + 31) - m] = 960 = 2^{6} \times 3 \times 5.$

Now if we choose divisor pairs $a$ and $b$ we will end up with the pair of equations

$(n + 31) + m = a, (n + 31) - m = b \Longrightarrow n = \dfrac{a + b}{2} - 31.$

This value is integral and maximized when $a + b$ is even and when $a$ and $b$ are as "far apart" as possible. The most "far apart" divisor pair is $(960,1),$ but the sum here is odd. The next pair to consider is $(480,2),$ which yields a potential maximal value of $n = 210.$ But first we need to check which values of $a$ result:

$a = \dfrac{635 \pm 239}{438} \Longrightarrow a = \dfrac{23}{11}$ or $a = \dfrac{18}{19}.$

Now looking back at the original sum, we see that the domain conditions of the log function require that $a \gt 2.$ Thus with this value of $n$ we have exactly one valid rational solution for $a,$ namely $a = \dfrac{23}{11},$ and so the desired value for $n$ is $\boxed{210}.$