(2) If under this condition, there is always $\sum_{i=1}^{n} x_{i} \geqslant \sqrt{\frac{n(n+1)}{2}}$, find all possible values of the positive integer $n$. (2008 China National Training Team Problem)
(2) All possible values of $n$ are $1,2,3$.

Sufficiency: When $n=\mathrm{F}$, from the condition $x_{1}^{2}=1$, we get $x_{1} \geqslant \sqrt{\frac{n(n+1)}{2}}$. For $n=2$, from the condition
$$x_{1}^{2}+x_{2}^{2}+x_{1}^{2} x_{2}^{2}=3$$

we need to prove $\square$
$$x_{1}+x_{2} \geqslant \sqrt{3}$$

Using proof by contradiction. If $x_{1}+x_{2}2 x_{1} x_{2}$, and $x_{1} x_{2}>0$, so it can only be that $x_{1} x_{2}>2$. In this case, by the AM-GM inequality, we have $x_{1}+x_{2} \geqslant 2 \sqrt{x_{1} x_{2}}>2 \sqrt{2}>\sqrt{3}$, which is a contradiction. Therefore, the proposition holds. For $n=3$, we have
$$x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{1}^{2} x_{2}^{2}+x_{1}^{2} x_{3}^{2}+x_{3}^{2} x_{2}^{2}=6$$

we need to prove $\square$
$$x_{1}+x_{2}+x_{3} \geqslant \sqrt{6}$$

Still using proof by contradiction. If $x_{1}+x_{2}+x_{3}2 x_{1} x_{2}+2 x_{1} x_{3}+2 x_{3} x_{2}$, this implies that at least one of $x_{1}^{2} x_{2}^{2}>2 x_{1} x_{2}$, $x_{1}^{2} x_{3}^{2}>2 x_{1} x_{3}$, $x_{3}^{2} x_{2}^{2}>2 x_{3} x_{2}$ holds.

Without loss of generality, assume $x_{1}^{2} x_{2}^{2}>2 x_{1} x_{2}$, then $x_{1} x_{2}>2, x_{1}+x_{2} \geqslant 2 \sqrt{x_{1} x_{2}}>2 \sqrt{2}>\sqrt{6}$, which is a contradiction. Therefore, the sufficiency is proven.

Necessity: Suppose $n \geqslant 4$. Let $x_{3}=x_{4}=\cdots=x_{n}=0$, the condition becomes
$$x_{1}^{2}+x_{2}^{2}+x_{1}^{2} x_{2}^{2}=\frac{n(n+1)}{2} \Leftrightarrow\left(x_{1}+x_{2}\right)^{2}+\left(x_{1}^{2} x_{2}^{2}-2 x_{1} x_{2}\right)=\frac{n(n+1)}{2}$$

Let $\varepsilon>0$ be undetermined. Let $x_{1}, x_{2}$ satisfy $\left\{\begin{array}{l}x_{1} x_{2}=2+\varepsilon \\ \left(x_{1}+x_{2}\right)^{2}=\frac{n(n+1)}{2}-2 \varepsilon-\varepsilon^{2}\end{array}\right.$.
If such $x_{1}, x_{2}$ exist, then the proposition does not hold. This only requires proving that there exists $\varepsilon$ satisfying the condition, i.e., $\square$
$$\frac{n(n+1)}{2}-2 \varepsilon-\varepsilon^{2} \geqslant 4(2+\varepsilon) \Leftrightarrow \varepsilon^{2}+6 \varepsilon \geqslant \frac{n(n+1)}{2}-8$$

Since $n \geqslant 4$, we have $\frac{n(n+1)}{2} \geqslant 10>8$, so we can always choose $\varepsilon>0$ such that the above inequality holds, thus there exist $x_{1}, x_{2}, \cdots, x_{n}$ satisfying the condition and
$$x_{1}+x_{2}+\cdots+x_{n}<\sqrt{\frac{n(n+1)}{2}}$$