$2 \cdot 61$ (1) For $n$ mutually independent numbers $x_{1}, x_{2}, \cdots, x_{n}$ each taking values 1, 0, or -1, find the minimum value of the sum of the pairwise products of these $n$ numbers.
(2) Find the minimum value of the sum of the pairwise products of $n$ numbers whose absolute values do not exceed 1.
[Solution] (1) The sum of all possible pairwise products of $x_{1}, x_{2}, \cdots, x_{n}$, denoted as $S$, can be written as
$$
S=\frac{1}{2}\left(\left(x_{1}+x_{2}+\cdots+x_{n}\right)^{2}-x_{1}^{2}-x_{2}^{2}-\cdots-x_{n}^{2}\right) .
$$

From this, we can derive that $S \geqslant-\frac{n}{2}$.
If $n$ is even, we can take half of the $x_{k}$ to be 1 and the other half to be -1, then we get
$$
S=-\frac{n}{2} .
$$

If $n$ is odd, considering that $S$ is an integer, we have
$$
S \geqslant-\frac{n-1}{2} \text {. }
$$

In this case, we can take $\frac{n+1}{2}$ of the $x_{k}$ to be 1 and $\frac{n-1}{2}$ of the $x_{k}$ to be -1, then we get
$$
S=-\frac{n-1}{2} \text {. }
$$
(2) This can be reduced to (1): Each $x_{k}$ can be sequentially replaced by 1 or -1, such that the sum of all possible pairwise products does not increase (replacement rule: if the sum of the remaining numbers is non-negative, then $x_{k}$ is changed to -1; if the sum of the remaining numbers is negative, then $x_{k}$ is 1). Therefore, we have the same answer as in (1).