Consider such a set $X \subseteq S(m, n)$. Clearly, $S(m, n)$ and thus a fortiori $X$ are finite sets. Then
$$\sum_{\mathbf{x} \in X} x_k \ge \sum_{i=1}^{|X|} (i-1) = |X|(|X| - 1)/2, \text{ for all } k = 1, 2, \dots, m,$$
and so
$$n|X| = \sum_{\mathbf{x} \in X} \sum_{k=1}^{m} x_k = \sum_{k=1}^{m} \sum_{\mathbf{x} \in X} x_k \ge m|X|(|X| - 1)/2,$$
hence $n \ge m(|X| - 1)/2$, therefore $|X| \le \left\lfloor \frac{2n}{m} \right\rfloor + 1$. This bound is thus valid for all points, and we claim it is sharp in all cases,¹¹ so $N(m, n) = \left\lfloor \frac{2n}{m} \right\rfloor + 1$.

Let us now, for $m = 3$, build a model set $X$ with $|X| = \left\lfloor \frac{2n}{3} \right\rfloor + 1$.
When $n = 3k-1$, the above expression evaluates at $2k$, and one can use the triplets $(0, k+1, 2k-2)$, $(1, k+2, 2k-4)$, ..., $(k-1, 2k, 0)$, $(k, 0, 2k-1)$, $(k+1, 1, 2k-3)$, ..., $(2k-1, k-1, 1)$.
When $n = 3k$, the above expression evaluates at $2k+1$, and one can use the triplets $(0, k, 2k)$, $(1, k+1, 2k-2)$, ..., $(k, 2k, 0)$, $(k+1, 0, 2k-1)$, $(k+2, 1, 2k-3)$, ..., $(2k, k-1, 1)$.
When $n = 3k + 1$, the above expression evaluates at $2k + 1$, and one can use the triplets $(0, k, 2k+1)$, $(1, k+1, 2k-1)$, ..., $(k, 2k, 1)$, $(k+1, 0, 2k)$, $(k+2, 1, 2k-2)$, ..., $(2k, k-1, 2)$.

Denote now, in the general case, $n' = \left\lfloor \frac{2n}{m} \right\rfloor$ and $m' = m - 2$. Then $N(2, n') = n' + 1$. We will describe how to build a model inductively.

¹¹$N(2, n) = n + 1 = |S(2, n)|$, as trivially seen, while $N(2n, n) = 2$, given by e.g. $(0, 0, \dots, 0, 1, 1, \dots, 1)$ and $(1, 1, \dots, 1, 0, 0, \dots, 0)$.

Now $\lfloor \frac{2(n-n')}{m'} \rfloor \ge \lfloor \frac{2n}{m} \rfloor$, since $\frac{2(n-n')}{m'} \ge \frac{2n}{m}$ is equivalent to $m(n-n') \ge (m-2)n$, or $2n \ge mn'$, or $n' \le \frac{2n}{m}$, patently true given the definition of $n'$, so the model for $N(m', n-n')$ yields enough elements that may be adjoined to those of the model realizing $N(2, n')$, in order to create one for $N(m, n)$ (valid, since $m = m' + 2$ and $n = (n-n') + n'$).

For the quite interesting particular case $m = n$, a model set $X = \{\mathbf{x}, \mathbf{y}, \mathbf{z}\}$ with $|X| = \lfloor \frac{2n}{n} \rfloor + 1 = 3$ is also easily built inductively.
Start with $n=2$ and $\mathbf{x} = (0,2), \mathbf{y} = (2,0), \mathbf{z} = (1,1)$. Have $\mathbf{x} = (x_1, \dots, x_k, 2)$, $\mathbf{y} = (y_1, \dots, y_k, 0)$, $\mathbf{z} = (z_1, \dots, z_k, 1)$ already built, for some $1 \le k < n-1$, and build $\mathbf{x} = (z_1, \dots, z_k, 0,2)$, $\mathbf{y} = (y_1, \dots, y_k, 1,0)$, $\mathbf{z} = (x_1, \dots, x_k, 2,1)$, until reaching the full $n$ coordinates.