Sure, here is the translated text:

```
II. (40 points) Find all positive integers $n$, such that for any positive real numbers $a, b, c$ satisfying $a+b+c=1$, we have
$$
a b c\left(a^{n}+b^{n}+c^{n}\right) \leqslant \frac{1}{3^{n+2}}.
$$
```
(1) When $n \geqslant 3$, take $a=\frac{2}{3}, b=c=\frac{1}{6}$.
Then $a b c\left(a^{n}+b^{n}+c^{n}\right)$
$$
=\frac{1}{3^{n+3}}\left(2^{n-1}+\frac{1}{2^{n}}+\frac{1}{2^{n}}\right)>\frac{1}{3^{n+2}} \text {. }
$$

Thus, $n \geqslant 3$ does not satisfy the condition.
(2) When $n=1$,
$$
a b c(a+b+c)=a b c \leqslant\left(\frac{a+b+c}{3}\right)^{3} \leqslant \frac{1}{3^{3}} \text {. }
$$

Thus, $n=1$ satisfies the condition.
(3) When $n=2$, the original inequality also holds.
$$
\begin{array}{l}
\text { Let } x=a b+b c+c a \text {. } \\
\text { Then, } a^{2}+b^{2}+c^{2}=1-2 x . \\
\text { By }(a b+b c+c a)^{2} \geqslant 3 a b c(a+b+c) \\
\Rightarrow 3 a b c \leqslant x^{2} \\
\Rightarrow a b c\left(a^{2}+b^{2}+c^{2}\right) \leqslant \frac{1}{3} x^{2}(1-2 x) \\
\Rightarrow 0<x<\frac{1}{2} \\
\Rightarrow \frac{1}{3} x^{2}(1-2 x) \leqslant \frac{1}{3}\left(\frac{x+x+1-2 x}{3}\right)^{3}=\frac{1}{3^{4}} \\
\Rightarrow a b c\left(a^{2}+b^{2}+c^{2}\right) \leqslant \frac{1}{3} x^{2}(1-2 x) \leqslant \frac{1}{3^{4}} .
\end{array}
$$