Question 218, Find the smallest positive integer $n$, such that there exists an $(n+1)$-term sequence $\mathrm{a}_{0}, \mathrm{a}_{1}, \ldots, a_{n}$, satisfying $a_{0}=0, a_{n}=2008$, and $\left|a_{i}-a_{i-1}\right|=i^{2}, i=1, 2, \ldots, n$.
Question 218, Solution: According to the condition, we have $a_{i}-a_{i-1}= \pm i^{2}$. Suppose there are exactly $k$ values of $i$ such that $a_{1}-a_{i-1}=-i^{2}$, which are $1 \leq i_{1}<i_{2}<\ldots<i_{k} \leq n$, then
$$
\begin{array}{l}
2008=a_{n}=\sum_{i=1}^{n}\left(a_{i}-a_{i-1}\right)=\sum_{i=1}^{n} i^{2}-2\left(i_{1}^{2}+i_{2}^{2}+\ldots+i_{k}^{2}\right) \\
\Rightarrow \frac{n(n+1)(2 n+1)}{6}=2008+2\left(i_{1}^{2}+i_{2}^{2}+\ldots+i_{k}^{2}\right)
\end{array}
$$

Thus, we know $\frac{n(n+1)(2 n+1)}{6} \geq 2008 \Rightarrow n \geq 18$, and $\frac{n(n+1)(2 n+1)}{6}$ is even, so $n \neq 18$, hence $n \geq 19$.
On the other hand, when $n=19$, $\sum_{i=1}^{19} i^{2}-2\left(1^{2}+3^{2}+10^{2}+11^{2}\right)=2016$, this way we can get a sequence that meets the condition.
In summary, the minimum value of $n$ that satisfies the condition is 19.