Five. (20 points) The sequence of positive integers $\left\{a_{n}\right\}$ satisfies:
$$
a_{1}=1, a_{n+1}=\left\{\begin{array}{l}
a_{n}-n, a_{n}>n ; \\
a_{n}+n, a_{n} \leqslant n .
\end{array}\right.
$$
(1) Find $a_{2008}$;
(2) Find the smallest positive integer $n$, such that $a_{n}=2008$.
Five, the initial values of the easy sequence (see Table 1).
Table 1
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}
\hline$n$ & 1 & $2 \mid 3$ & & & 6 & 7 & 8 & & 10 & 11 & 12 & & \\
\hline$a$ & 1 & 2 & 41 & & 10 & & 11 & & $12 \mid$ & 2 & 13 & 1 & 14 \\
\hline
\end{tabular}

Next, focus on the subscripts $n_{k}$ such that $a_{n_{k}}=1$: $n_{1}=1$, $n_{2}=4, n_{3}=13, \cdots \cdots$ They satisfy the following recursive relation:
$$
n_{k+1}=3 n_{k}+1(k=1,2, \cdots) \text {. }
$$

We will use induction on $k$.
When $k=1,2$, equation (1) holds.
Assume $a_{n_{k}}=1$, then by the condition
$$
\begin{array}{l}
a_{n_{k}+1}=n_{k}+1, a_{n_{k}+2}=2 n_{k}+2, \\
a_{n_{k}+3}=n_{k}, a_{n_{k}+4}=2 n_{k}+3, \cdots \cdots
\end{array}
$$

By induction, we get
$$
\begin{array}{c}
a_{n_{k}+2 m-1}=n_{k}+2-m\left(m=1,2, \cdots, n_{k}+1\right), \\
a_{n_{k}+2 m}=2 n_{k}+1+m\left(m=1,2, \cdots, n_{k}\right) .
\end{array}
$$

Thus, when $m=n_{k}+1$,
$$
a_{3 n_{k}+1}=n_{k}+2-\left(n_{k}+1\right)=1 \text {. }
$$

Therefore, $n_{k+1}=3 n_{k}+1(k=1,2, \cdots)$, i.e., equation (1) holds.
From equation (1), we get $2 n_{k+1}+1=3\left(2 n_{k}+1\right)$.
Let $2 n_{k}+1=x_{k}$. Then $x_{k+1}=3 x_{k}, x_{1}=3$.
So, $x_{k}=3^{k}$.
Therefore, $n_{k}=\frac{3^{k}-1}{2}(k=1,2, \cdots)$.
And $n_{7}=\frac{3^{7}-1}{2}=1093, n_{8}=\frac{3^{8}-1}{2}=3280$, thus $n_{7}<2008<n_{8}$.
Also, $2008=n_{7}+2 \times 458-1$, so by equation (2), $a_{2008}=n_{7}+2-458=637$.
From equation (2), we know that when $n \leqslant 3 n_{k}=n_{k+1}-1$, $a_{n} \leqslant 3 n_{k}+1=n_{k+1}$.
Therefore, when $n<n_{7}$, $a_{n} \leqslant n_{7}=1093$.
And when $n_{7} \leqslant n<n_{8}$, either $a_{n} \leqslant 1094$, or $a_{n} \geqslant 2 \times 1094$, i.e., $a_{n}$ cannot take the value 2008.
Furthermore, consider the case $n_{8} \leqslant n<n_{9}$.
From $n_{8}+2-m=2008$, we get $m=1274$.
By equation (2),
$a_{5827}=a_{n_{8}+2 m-1}=n_{8}+2-m=2008$.
Thus, the smallest $n$ such that $a_{n}=2008$ is 5827.