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

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

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

When $n = 3m$, for $1 \leq j \leq m+1$, let $x_j = j-1$, $y_j = m + j - 1$, $z_j = 2m - 2j + 2$; for $m + 2 \leq j \leq 2m + 1$, 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 $1 \leq j \leq m$, let $x_j = j - 1$, $y_j = m + j$, $z_j = 2m - 2j + 2$; for $m + 1 \leq j \leq 2m$, 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 $1 \leq j \leq m + 1$, let $x_j = j - 1$, $y_j = m + j$, $z_j = 2m - 2j + 3$; for $m + 2 \leq j \leq 2m + 1$, 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 $\lfloor \frac{2n}{3} \rfloor + 1$.