China Western Mathematical Olympiad

By the given condition, we have $$kn\ge \sum _{i=1}^k(x_i+y_i+z_i)\ge 3\sum _{i=0}^{k-1}i=\frac{3k(k-1)}{2},$$ and then $k\le \lfloor \frac{2n}{3}\rfloor +1$.

The following illustrates the case of $k=\lfloor \frac{2n}{3}\rfloor +1$:

Set $m\in {\mathbb{Z}}^{+}$.

When $n=3m$, for $1\le j\le m+1$, let $x_j=j-1$, $y_j=m+j-1$, $z_j=2m-2j+2$; for $m+2\le j\le 2m+1$, let $x_j=j-1$, $y_j=j-m-2$, $z_j=4m-2j+3$, and the result is obvious.

When $n=3m+1$, for $1\le j\le m$, let $x_j=j-1$, $y_j=m+j$, $z_j=2m-2j+2$; for $m+1\le j\le 2m$, let $x_j=j+1$, $y_j=j-m-1$, $z_j=4m+1-2j$; and $x_{2m+1}=m$, $y_{2m+1}=2m+1$, $z_{2m+1}=0$ will lead to the expected result.

When $n=3m+2$, for $1\le j\le m+1$, let $x_j=j-1$, $y_j=m+j$, $z_j=2m-2j+3$; for $m+2\le j\le 2m+1$, let $x_j=j$, $y_j=j-m-2$, $z_j=4m-2j+4$; and $x_{2m+2}=2m+2$, $y_{2m+2}=m$, $z_{2m+2}=0$, and the result follows.

In summary, the maximum value of $k$ is $\lfloor \frac{2n}{3}\rfloor +1$.