The 63rd question: For positive integers $n \geq 2$, for complex numbers $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{n}$ satisfying $\left|\alpha_{i}\right| \leq 1$ (i=1, 2, \ldots, n), find the minimum possible value of $\sum_{\mathrm{i}=1}^{\mathrm{n}}\left|1+\alpha_{i}\right|+\left|1+\prod_{\mathrm{i}=1}^{\mathrm{n}} \alpha_{i}\right|$.
Question 63:
Solution: (1) When $n$ is odd, take $\alpha_{i}=-1(1 \leq i \leq n)$, then $\sum_{i=1}^{n}\left|1+\alpha_{i}\right|+\mid 1+$ $\prod_{i=1}^{n} \alpha_{i} \mid=0$. Clearly, $\sum_{i=1}^{n}\left|1+\alpha_{i}\right|+\left|1+\prod_{i=1}^{n} \alpha_{i}\right| \geq 0$, so the minimum value of $\sum_{i=1}^{n}\left|1+\alpha_{i}\right|+\left|1+\prod_{i=1}^{n} \alpha_{i}\right|$ at this time is 0.
(2) When $n$ is even, take $\alpha_{1}=\alpha_{2}=\ldots=\alpha_{n-1}=-1, \alpha_{n}=1$, then
$$
\sum_{\mathrm{i}=1}^{\mathrm{n}}\left|1+\alpha_{\mathrm{i}}\right|+\left|1+\prod_{\mathrm{i}=1}^{\mathrm{n}} \alpha_{i}\right|=2
$$

We will use induction on even $n$ to prove: $\sum_{\mathrm{i}=1}^{\mathrm{n}}\left|1+\alpha_{i}\right|+\left|1+\prod_{\mathrm{i}=1}^{\mathrm{n}} \alpha_{i}\right| \geq 2$.
First, we prove a lemma: If the modulus of complex numbers $\alpha, \beta, \gamma$ are all no greater than 1, then $|1+\alpha|+|1+\beta|+|1+\gamma| \geq$ $|1+\alpha \beta \gamma|$.
Proof of the lemma: In fact,
$$
\begin{array}{l}
|1+\alpha|+|1+\beta|+|1+\gamma| \\
\geq|1+\alpha|+|1+\beta| \cdot|\alpha|+|1+\gamma| \cdot|\alpha \beta| \\
\geq|(1+\alpha)-(1+\beta) \alpha+(1+\gamma) \alpha \beta|=|1+\alpha \beta \gamma|
\end{array}
$$

The lemma is proved.
Now, let's return to the original problem.
When $\mathrm{n}=2$,
$$
\begin{array}{l}
\left|1+\alpha_{1}\right|+\left|1+\alpha_{2}\right|+\left|1+\alpha_{1} \alpha_{2}\right| \\
\geq\left|1+\alpha_{1}\right|+\left|1+\alpha_{2}\right| \cdot\left|\alpha_{1}\right|+\left|1+\alpha_{1} \alpha_{2}\right| \\
\geq\left|\left(1+\alpha_{1}\right)-\left(1+\alpha_{2}\right) \alpha_{1}+1+\alpha_{1} \alpha_{2}\right|=2
\end{array}
$$

The conclusion holds.
Assume that the conclusion holds for $\mathrm{n}$. Then for $\mathrm{n}+2$, according to the lemma and the induction hypothesis, we have:
$$
\begin{array}{l}
\sum_{i=1}^{n+2}\left|1+\alpha_{i}\right|+\left|1+\prod_{i=1}^{n+2} \alpha_{i}\right| \\
=\sum_{i=1}^{n-1}\left|1+\alpha_{i}\right|+\left|1+\alpha_{n}\right|+\left|1+\alpha_{n+1}\right|+\left|1+\alpha_{n+2}\right|+\left|1+\prod_{i=1}^{n-1} \alpha_{i} \cdot\left(\alpha_{n} \alpha_{n+1} \alpha_{n+2}\right)\right| \\
\geq \sum_{i=1}^{n-1}\left|1+\alpha_{i}\right|+\left|1+\alpha_{n} \alpha_{n+1} \alpha_{n+2}\right|+\left|1+\prod_{i=1}^{n-1} \alpha_{i} \cdot\left(\alpha_{n} \alpha_{n+1} \alpha_{n+2}\right)\right| \geq 2
\end{array}
$$

By mathematical induction, the conclusion holds.
In summary, the minimum possible value of $\sum_{\mathrm{i}=1}^{\mathrm{n}}\left|1+\alpha_{\mathrm{i}}\right|+\left|1+\prod_{\mathrm{i}=1}^{\mathrm{n}} \alpha_{\mathrm{i}}\right|$ is $1+(-1)^{\mathrm{n}}$.