Solution:

For $n \ge 3,$ both $a_{n+1}a_{n-1} = a_n^2 + 2007$ and $a_n a_{n-2} = a_{n-1}^2 + 2007$ hold. Subtracting and regrouping gives $a_{n-1}(a_{n+1} + a_{n-1}) = a_n(a_n + a_{n-2}),$ so $\frac{a_{n+1} + a_{n-1}}{a_n}$ has the same value for every $n \ge 2.$ Since $a_3 = \frac{3^2 + 2007}{3} = 672,$ that value is $\frac{672 + 3}{3} = 225,$ and the sequence satisfies $a_{n+1} = 225a_n - a_{n-1}.$

Multiplying $a_{n+1} + a_{n-1} = 225a_n$ by $a_{n+1}$ and substituting $a_{n+1}a_{n-1} = a_n^2 + 2007$ yields $a_{n+1}^2 + a_n^2 + 2007 = 225\,a_n a_{n+1},$ so $$\frac{a_{n+1}^2 + a_n^2}{a_{n+1}a_n} = 225 - \frac{2007}{a_n a_{n+1}}.$$

The sequence increases: $a_3 = 672 \gt a_2,$ and $a_{n+1} = 225a_n - a_{n-1} \gt a_n$ whenever $a_n \gt a_{n-1}.$ Hence $a_{2006}a_{2007} \gt 672^2 \gt 2007,$ so the fraction lies strictly between $224$ and $225,$ and the answer is $224.$