4.6.43 $\stackrel{\star \star}{\star}$ Let $x_{1}, x_{2}, \cdots, x_{n}$ be non-negative real numbers, and satisfy
$$
\sum_{i=1}^{n} x_{i}^{2}+\sum_{1 \leqslant i<j \leqslant n}\left(x_{i} x_{j}\right)^{2}=\frac{n(n+1)}{2} \text {. }
$$
(1) Find the maximum value of $\sum_{i=1}^{n} x_{i}$;
(2) Find all positive integers $n$, such that $\sum_{i=1}^{n} x_{i} \geqslant \sqrt{\frac{n(n+1)}{2}}$.
(1) $\left(\sum_{i=1}^{n} x_{i}\right)^{2}=\sum_{i=1}^{n} x_{i}^{2}+2 \sum_{12}$, suppose $x_{1} x_{2}>2$, then
$$
x_{1}+x_{2}+x_{3} \geqslant x_{1}+x_{2} \geqslant 2 \sqrt{x_{1} x_{2}}>2 \sqrt{2}>\sqrt{6} \text {. }
$$

When $n \geqslant 4$, let $x_{1}=x_{2}=x, x_{3}=\cdots=x_{n}=0$, then $2 x^{2}+x^{4}=\frac{n(n+1)}{2}$, solving for $x$ gives $x=\sqrt{\frac{\sqrt{2 n(n+1)+4}-2}{2}}$, at this point, $\sum_{i=1}^{n} x_{i}=2 \sqrt{\frac{\sqrt{2 n(n+1)+4}-2}{2}}16$, where $t=n(n+1)$.
In summary, the positive integers sought are $n=1,2,3$.