6. For any positive integers $x, y, z$, if $x \mid y^{3}$, $y \mid z^{3}$, $z \mid x^{3}$, and $x y z \mid (x+y+z)^{n}$, find the smallest positive integer $n$ that satisfies the condition.
6. The minimum value of $n$ is 13.

Notice that $x y z \mid (x+y+z)^{n}$ if and only if for every prime $p$, we have
$$
v_{p}(x y z) \leqslant v_{p}\left((x+y+z)^{n}\right).
$$

Let $p$ be any prime, and assume without loss of generality that
$$
v_{p}(z)=\min \left\{v_{p}(x), v_{p}(y), v_{p}(z)\right\}.
$$

Then, $v_{p}(x+y+z) \geqslant v_{p}(z)$.
From $x \mid y^{3}, y \mid z^{3}, z \mid x^{3}$, we know
$$
v_{p}(x) \leqslant 3 v_{p}(y) \leqslant 9 v_{p}(z).
$$
$$
\begin{array}{l}
\text{Then } v_{p}(x y z)=v_{p}(x)+v_{p}(y)+v_{p}(z) \\
\leqslant 9 v_{p}(z)+3 v_{p}(z)+v_{p}(z)=13 v_{p}(z) \\
\leqslant 13 v_{p}(x+y+z) \\
=v_{p}\left((x+y+z)^{13}\right).
\end{array}
$$

This shows that $n=13$ satisfies the problem's requirements.
Next, we prove that $n=13$ is the minimum value.
Let $n$ be a positive integer that satisfies the problem's requirements. By
$$
(x, y, z)=\left(p^{9}, p^{3}, p^{1}\right)
$$

we get that for any prime $p$,
$$
\begin{array}{l}
13=v_{p}\left(p^{13}\right)=v_{p}\left(p^{9} p^{3} p^{1}\right)=v_{p}(x y z) \\
\leqslant v_{p}\left((x+y+z)^{n}\right)=v_{p}\left(\left(p^{9}+p^{3}+p^{1}\right)^{n}\right) \\
=n v_{p}\left(p\left(p^{8}+p^{2}+1\right)\right)=n.
\end{array}
$$

Therefore, $n \geqslant 13$.
In conclusion, the minimum value of $n$ that satisfies the problem's requirements is 13.