Let $n \geq 2$ be an integer and let $\{T_{1}, \ldots, T_{N}\}$ be any set of triples of nonnegative integers satisfying the conditions (1) and (2). Since the $a$-coordinates are pairwise distinct we have
$$\sum_{i=1}^{N} a_{i} \geq \sum_{i=1}^{N}(i-1)=\frac{N(N-1)}{2}$$
Analogously,
$$\sum_{i=1}^{N} b_{i} \geq \frac{N(N-1)}{2} \quad \text{ and } \quad \sum_{i=1}^{N} c_{i} \geq \frac{N(N-1)}{2} .$$
Summing these three inequalities and applying (1) yields
$$3 \frac{N(N-1)}{2} \leq \sum_{i=1}^{N} a_{i}+\sum_{i=1}^{N} b_{i}+\sum_{i=1}^{N} c_{i}=\sum_{i=1}^{N}\left(a_{i}+b_{i}+c_{i}\right)=n N,$$
hence $3 \frac{N-1}{2} \leq n$ and, consequently,
$$N \leq\left\lfloor\frac{2 n}{3}\right\rfloor+1$$
By constructing examples, we show that this upper bound can be attained, so $N(n)=\left\lfloor\frac{2 n}{3}\right\rfloor+1$.
We distinguish the cases $n=3 k-1, n=3 k$ and $n=3 k+1$ for $k \geq 1$ and present the extremal examples in form of a table.