Is there a quicker way?

For each positive integer $n$ ,

$\frac{n^2}{n^2-24n+288} + \frac{(24-n)^2}{(24-n)^2-24(24-n)+288}$

$=\frac{2n^2}{2n^2-48n+576} + \frac{2(24-n)^2}{2(24-n)^2-48(24-n)+576}$

$=\frac{2n^2}{n^2+(24-n)^2} + \frac{2(24-n)^2}{(24-n)^2+n^2} = 2$

Then,

$\frac{1^2}{1^2-24+288} + \frac{2^2}{2^2-48+288} + ... + \frac{23^2}{23^2-23(24)+288} + \frac{24^2}{24^2-24(24)+288}$

$= (\frac{1^2}{1^2-24+288} + \frac{23^2}{23^2-23(24)+288}) + (\frac{2^2}{2^2-48+288} + \frac{22^2}{22^2-22(24)+288}) + ... + ( \frac{11^2}{11^2-11(24)+288} + \frac{13^2}{13^2-13(24)+288}) + \frac{12^2}{12^2-12(24)+288} + \frac{24^2}{24^2-24(24)+288}$

$=11(2) + 1 + 2$

$=\boxed{25}$

The $n$ th term $a_n = \dfrac {n^2}{n^2 - 24n + 288} = \dfrac {n^2}{(n - 12)^2 + 144}$

For $n<12$ , adding $n$ th and $(24-n)$ th terms together:

$\begin{aligned} a_n + a_{24-n} & = \frac {n^2}{(12-n)^2 + 144} + \frac {(24-n)^2}{(24-n-12)^2 + 144} \\ & = \frac {n^2}{(12-n)^2 + 144} + \frac {n^2 - 48n + 576}{(12-n)^2 + 144} \\ & = \frac {2(n^2-24n+288)}{n^2 - 24n + 288} \\ & = 2 \end{aligned}$

$\begin{aligned} S & = a_1+a_2+a_3 + ... + a_{24} \\ & = \color{#3D99F6}{(a_1+a_{23}) + (a_2+a_{22}) + ... + (a_{11}+a_{13})} + a_{12} + a_{24} \\ & = \color{#3D99F6}{11(2)} + \frac {12^2}{0^2+144} + \frac {24^2}{12^2+144} \\ & = 22 + 1 + 2 \\ & = \boxed{25} \end{aligned}$