It's easy to see that if $a_n$ is rational, then $a_{n+1}$ is also rational, but since $a_1$ is rational, it follows that $a_n$ is rational for all $n \geq 1$. Let $a_n = \frac{p_n}{q_n}$, where $\{ p_n \}_{n\geq 1}$ and $\{ q_n \}_{n \geq 1}$ are two sequences of integers, then \[ a_{n+1} = \frac{n}{\frac{p_n}{q_n} + 1} \\ \\ \frac{p_{n+1}}{q_{n+1}} = \frac{n q_n}{p_n + q_n} \\ \\ \implies p_{n+1} = n q_n \ \text{and} \ q_{n+1} = p_n+q_n \\ \\ \implies \frac{p_{n+2}}{n+1} = p_n + \frac{p_{n+1}}{n} \text{and} \ q_{n+1} = (n-1)q_{n-1} + q_n \\ \\ \implies p_{n+2} = \frac{n+1}{n} p_{n+1} + (n+1)p_n \ \text{and} \ q_{n+2} = q_{n+1} + nq_n \\ \] Now, just find $p_n$ and $q_n$ and use it to find $a_{2008}$.