For any integer $n\ge 2$, let $N(n)$ be the maximum number of triples $(a_j,b_j,c_j),j=1,2,3,\cdots ,N(n),$ consisting of non-negative integers $a_j,b_j,c_j$ (not necessarily distinct) such that the following two conditions are satisfied:

(a) $a_j+b_j+c_j=n,$ for all $j=1,2,3,\cdots N(n)$;
(b) $j\neq k$, then $a_j\neq a_k$, $b_j\neq b_k$ and $c_j\neq c_k$.

Determine $N(n)$ for all $n\ge 2$.
1. **Understanding the Problem:**
   We need to find the maximum number of triples \((a_j, b_j, c_j)\) consisting of non-negative integers such that:
   - \(a_j + b_j + c_j = n\) for all \(j\).
   - If \(j \neq k\), then \(a_j \neq a_k\), \(b_j \neq b_k\), and \(c_j \neq c_k\).

2. **Summing the Elements:**
   Since all numbers in the set \(\{a_i\}_{i=1}^{N(n)}\) are distinct and non-negative integers, we have:
   \[
   \sum_{i=1}^{N(n)} a_i \geq 0 + 1 + \cdots + (N(n)-1) = \frac{N(n) (N(n)-1)}{2}.
   \]
   Similarly, for \(\{b_i\}_{i=1}^{N(n)}\) and \(\{c_i\}_{i=1}^{N(n)}\), we get:
   \[
   \sum_{i=1}^{N(n)} b_i \geq \frac{N(n)(N(n)-1)}{2},
   \]
   \[
   \sum_{i=1}^{N(n)} c_i \geq \frac{N(n)(N(n)-1)}{2}.
   \]

3. **Adding the Inequalities:**
   Adding these inequalities, we obtain:
   \[
   \sum_{i=1}^{N(n)} a_i + b_i + c_i \geq \frac{3}{2} N(n) (N(n)-1).
   \]
   But, we know that:
   \[
   \sum_{i=1}^{N(n)} (a_i + b_i + c_i) = N(n) \cdot n.
   \]
   Therefore, we have:
   \[
   \frac{3}{2} N(n) (N(n)-1) \leq N(n) \cdot n.
   \]

4. **Solving the Inequality:**
   Dividing both sides by \(N(n)\), we get:
   \[
   \frac{3}{2} (N(n) - 1) \leq n.
   \]
   Solving for \(N(n)\), we find:
   \[
   N(n) - 1 \leq \frac{2n}{3},
   \]
   \[
   N(n) \leq \frac{2n}{3} + 1.
   \]
   Since \(N(n)\) must be an integer, we take the floor function:
   \[
   N(n) \leq \left\lfloor \frac{2n}{3} + 1 \right\rfloor.
   \]

5. **Constructing the Triples:**
   To show that this bound is achievable, we do casework on \(n \pmod{3}\):
   - For \(n = 3k\), we can set \((a_i, b_i, c_i) = (i, k+i, 2k - i)\) for \(i = 0, 1, \ldots, k-1\).
   - For \(n = 3k+1\), we can set \((a_i, b_i, c_i) = (i, k+i, 2k + 1 - 2i)\) for \(i = 0, 1, \ldots, k\).
   - For \(n = 3k+2\), we can set \((a_i, b_i, c_i) = (i, k+i, 2k + 2 - 2i)\) for \(i = 0, 1, \ldots, k\).

   In each case, the number of triples is \(\left\lfloor \frac{2n}{3} + 1 \right\rfloor\).

The final answer is \(\boxed{\left\lfloor \frac{2n}{3} + 1 \right\rfloor}\).