1. Find the smallest positive integer $n$, such that the set $\{1,2,3, \cdots, 3 n-1,3 n\}$ can be divided into $n$ mutually disjoint triples $\{x, y, z\}$, where $x+y=3 z$.
1. Let $n$ be a positive integer with the property mentioned in the problem, and let the $n$ triples be $\left\{x_{i}, y_{i}, z_{i}\right\}, i=1,2, \cdots, n$. Then, from $x_{i}+y_{i}=3 z_{i}$, we have $4 \sum_{i=1}^{n} z_{i}=\sum_{i=1}^{n}\left(x_{i}+y_{i}+z_{i}\right)=\sum_{j=1}^{3 n} j=\frac{3 n(3 n+1)}{2}$. Therefore, $8 \mid 3 n(3 n+1)$. Noting that $3 n$ and $3 n+1$ have different parities, we have $8 \mid 3 n$ or $8 \mid 3 n+1$, thus $n \geqslant 5$.
When $n=5$, there exists a partition of $\{1,2, \cdots, 15\}$ that satisfies the condition, for example:
$$
\{1,11,4\},\{2,13,5\},\{3,15,6\},\{9,12,7\},\{10,14,8\} \text {. }
$$

Therefore, the minimum value sought is 5.