The fact that the equation $ a_{n+1}a_{n-1} = a_n^2 + 2007 $ holds for $ n \geq 2 $ implies that $ a_na_{n-2} = a_{n-1}^2 + 2007 $ for $ n \geq 3 $. Subtracting the second equation from the first one yields $ a_{n+1} a_{n-1} -a_n a_{n-2} = a_n^2 -a_{n-1}^2 $, or \[a_{n+1} a_{n-1} + a_{n-1}^2 = a_n a_{n-2} + a_n^2.\]Dividing the last equation by $ a_{n-1} a_n $ and simplifying produces \[\frac{a_{n+1}+ a_{n-1}}{a_n}=\frac{a_n+a_{n-2}}{a_{n-1}}.\]This equation shows that $ \frac{a_{n+1}+a_{n-1}}{a_n} $ is constant for $ n\geq 2 $. Because $ a_3a_1 = a_2^2 + 2007 $, $ a_3=2016/3=672 $. Thus \[\frac{a_{n+1}+a_{n-1}}{a_n} = \frac{672+3}{3}=225,\]and $ a_{n+1}=225a_n-a_{n-1} $ for $ n \geq 2 $. Note that $ a_3 = 672 >3 = a_2 $. Furthermore, if $ a_n > a_{n-1} $, then $ a_{n+1}a_{n-1} = a_n^2 + 2007 $ implies that \[a_{n+1} = \frac{a_n^2}{a_{n-1}}+\frac{2007}{a_{n-1}} = a_n\left(\frac{a_n}{a_{n-1}}\right) + \frac{2007}{a_{n-1}}>a_n + \frac{2007}{a_{n-1}} > a_n.\]Thus by mathematical induction, $ a_n > a_{n-1} $ for all $ n \geq 3 $. Therefore the recurrence $ a_{n+1} = 225a_n - a_{n-1} $ implies that $ a_{n+1}> 225a_n - a_n = 224a_n $ and therefore $ a_n \geq 2007 $ for $ n \geq 4 $. Finding $ a_{n+1} $ from $ a_{n+1} a_{n-1} = a_n^2+ 2007 $ and substituting into $ 225 = \frac{a_{n+1}+a_{n-1}}{a_n} $ shows that \[\frac{a_n^2 + a_{n-1}^2}{a_n a_{n-1}} = 225 -\frac{2007}{a_n a_{n-1}}.\]Thus the largest integer less than or equal to the original fraction is $ \boxed{224} $.