The sequence ${a_0, a_1, a_2, ...}$ of real numbers satisfies the recursive relation $$n(n+1)a_{n+1}+(n-2)a_{n-1} = n(n-1)a_n$$ for every positive integer $n$, where $a_0 = a_1 = 1$. Calculate the sum $$\frac{a_0}{a_1} + \frac{a_1}{a_2} + ... + \frac{a_{2008}}{a_{2009}}$$.
1. **Identify the recursive relation and initial conditions:**
   The sequence \(\{a_n\}\) satisfies the recursive relation:
   \[
   n(n+1)a_{n+1} + (n-2)a_{n-1} = n(n-1)a_n
   \]
   with initial conditions \(a_0 = 1\) and \(a_1 = 1\).

2. **Assume a form for \(a_n\):**
   Let's assume \(a_n = \frac{1}{n!}\). We will use mathematical induction to verify this assumption.

3. **Base cases:**
   For \(n = 0\):
   \[
   a_0 = \frac{1}{0!} = 1
   \]
   For \(n = 1\):
   \[
   a_1 = \frac{1}{1!} = 1
   \]
   Both base cases match the given initial conditions.

4. **Inductive step:**
   Assume \(a_k = \frac{1}{k!}\) holds for \(k = n\) and \(k = n-1\). We need to show it holds for \(k = n+1\).

   Substitute \(a_n = \frac{1}{n!}\) and \(a_{n-1} = \frac{1}{(n-1)!}\) into the recursive relation:
   \[
   n(n+1)a_{n+1} + (n-2)a_{n-1} = n(n-1)a_n
   \]
   becomes
   \[
   n(n+1)a_{n+1} + (n-2)\frac{1}{(n-1)!} = n(n-1)\frac{1}{n!}
   \]
   Simplify the right-hand side:
   \[
   n(n-1)\frac{1}{n!} = \frac{n(n-1)}{n!} = \frac{n(n-1)}{n \cdot (n-1)!} = \frac{1}{(n-2)!}
   \]
   Now, simplify the left-hand side:
   \[
   n(n+1)a_{n+1} + \frac{(n-2)}{(n-1)!}
   \]
   To match the right-hand side, we need:
   \[
   n(n+1)a_{n+1} = \frac{1}{(n-2)!} - \frac{(n-2)}{(n-1)!}
   \]
   Simplify the right-hand side:
   \[
   \frac{1}{(n-2)!} - \frac{(n-2)}{(n-1)!} = \frac{(n-1) - (n-2)}{(n-1)!} = \frac{1}{(n-1)!}
   \]
   Thus,
   \[
   n(n+1)a_{n+1} = \frac{1}{(n-1)!}
   \]
   Therefore,
   \[
   a_{n+1} = \frac{1}{(n+1)!}
   \]
   This completes the induction, and we have \(a_n = \frac{1}{n!}\) for all \(n \geq 0\).

5. **Calculate the sum:**
   \[
   \sum_{k=0}^{2008} \frac{a_k}{a_{k+1}} = \sum_{k=0}^{2008} \frac{\frac{1}{k!}}{\frac{1}{(k+1)!}} = \sum_{k=0}^{2008} (k+1)
   \]
   This is an arithmetic series sum:
   \[
   \sum_{k=0}^{2008} (k+1) = 1 + 2 + 3 + \cdots + 2009
   \]
   The sum of the first \(n\) positive integers is given by:
   \[
   \sum_{k=1}^{n} k = \frac{n(n+1)}{2}
   \]
   Here, \(n = 2009\):
   \[
   \sum_{k=1}^{2009} k = \frac{2009 \cdot 2010}{2} = 2009 \cdot 1005
   \]

The final answer is \(\boxed{2009 \times 1005}\)