Three. (This question is worth 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 $\left|a_{i}-a_{i-1}\right|=i^{2}$, $i=1,2, \cdots, n$.

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If $n \leqslant 17$, then
$$
\begin{aligned}
a_{n} & =\sum_{i=1}^{n}\left(a_{i}-a_{i-1}\right)+a_{0} \\
& \leqslant \sum_{i=1}^{n}\left|a_{i}-a_{i-1}\right| \\
& =\frac{1}{6} n(n+1)(2 n+1) \\
& \leqslant \frac{1}{6} \times 17 \times 18 \times 35 \\
& <2008,
\end{aligned}
$$

which contradicts $a_{n}=2008$.
If $n=18$, then $a_{n}=\sum_{i=1}^{n}\left(a_{i}-a_{i-1}\right)+a_{0} \equiv \sum_{i=1}^{n}\left|a_{i}-a_{i-1}\right| \equiv \sum_{i=1}^{n} i^{2} \equiv$ $1(\bmod 2)$, which contradicts $a_{n}=2008$.
If $n=19$, note that
$$
2008=1^{2}+2^{2}+\cdots+19^{2}-2\left(2^{2}+5^{2}+9^{2}+11^{2}\right),
$$

take $a_{0}, a_{1}, \cdots, a_{19}$ as follows: $0,1,-3,6,22,-3,33,82,146,65,165$, $44,188,357,553,778,1034,1323,1647,2008$. This shows that $n=19$ is feasible.
In summary, $n_{\text {min }}=19$.
Note that the example is not unique, such as
$$
\begin{aligned}
2008 & =1^{2}+2^{2}+\cdots+19^{2}-2\left(1^{2}+3^{2}+10^{2}+11^{2}\right) \\
& =1^{2}+2^{2}+\cdots+19^{2}-2\left(2^{2}+3^{2}+4^{2}+9^{2}+11^{2}\right) \\
& =1^{2}+2^{2}+\cdots+19^{2}-2\left(1^{2}+3^{2}+5^{2}+14^{2}\right) \\
& =1^{2}+2^{2}+\cdots+19^{2}-2\left(2^{2}+3^{2}+7^{2}+13^{2}\right) .
\end{aligned}
$$