Three, (50 points) Find the smallest positive integer $n$ such that there exists an $(n+1)$-term sequence $a_{0}, a_{1}, \cdots, a_{n}$, satisfying $a_{0}=0, a_{n}=2008$, and
$$
\left|a_{i}-a_{i-1}\right|=i^{2}(i=1,2, \cdots, n) .
$$
If $n \leqslant 17$, then
$$
\begin{array}{l}
a_{n}=\sum_{i=1}^{n}\left(a_{i}-a_{i-1}\right)+a_{0} \\
\leqslant \sum_{i=1}^{n}\left|a_{i}-a_{i-1}\right|=\frac{1}{6} n(n+1)(2 n+1) \\
\leqslant \frac{1}{6} \times 17 \times 18 \times 35<2008,
\end{array}
$$

a contradiction.
If $n=18$, then
$$
\begin{array}{l}
a_{n}=\sum_{i=1}^{n}\left(a_{i}-a_{i-1}\right)+a_{0} \\
\equiv \sum_{i=1}^{n}\left|a_{i}-a_{i-1}\right| \equiv \sum_{i=1}^{n} i^{2} \equiv 1(\bmod 2),
\end{array}
$$

which contradicts $a_{n}=2008$.
If $n=19$, note that
$$
2008=1^{2}+2^{2}+\cdots+19^{2}-\left(2^{2}+5^{2}+9^{2}+11^{2}\right) \text {. }
$$

Take $a_{0}, a_{1}, \cdots, a_{19}$ as follows:
$$
\begin{array}{l}
0,1,-3,6,22,-3,33,82,146,65,165, \\
44,188,357,553,778,1034,1323, \\
1647,2008 .
\end{array}
$$

This shows that $n=19$ satisfies the requirement.
In summary, $n_{\min }=19$ satisfies the requirement.
Note: The example is not unique. For instance,
$$
2008=1^{2}+2^{2}+\cdots+19^{2}-2\left(1^{2}+3^{2}+10^{2}+11^{2}\right)
$$

provides another set of numbers.