For a given positive integer $n$, find the greatest positive integer $k$, such that there exist three sets of $k$ non-negative distinct integers, $A=\{x_1,x_2,\cdots,x_k\}, B=\{y_1,y_2,\cdots,y_k\}$ and $C=\{z_1,z_2,\cdots,z_k\}$ with $ x_j\plus{}y_j\plus{}z_j\equal{}n$ for any $ 1\leq j\leq k$.

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To find the greatest positive integer \( k \) such that there exist three sets of \( k \) non-negative distinct integers \( A = \{x_1, x_2, \cdots, x_k\} \), \( B = \{y_1, y_2, \cdots, y_k\} \), and \( C = \{z_1, z_2, \cdots, z_k\} \) with \( x_j + y_j + z_j = n \) for any \( 1 \leq j \leq k \), we can proceed as follows:

1. **Sum of Elements in Sets**:
   Define the sum of elements in set \( A \) as \( S_A \), in set \( B \) as \( S_B \), and in set \( C \) as \( S_C \). Then, the total sum of all elements in the sets \( A \), \( B \), and \( C \) is:
   \[
   S_A + S_B + S_C = kn
   \]

2. **Minimum Sum of Distinct Non-Negative Integers**:
   The minimum sum of \( k \) distinct non-negative integers is given by the sum of the first \( k \) non-negative integers:
   \[
   S_A \geq \frac{k(k-1)}{2}, \quad S_B \geq \frac{k(k-1)}{2}, \quad S_C \geq \frac{k(k-1)}{2}
   \]
   Therefore, the total minimum sum is:
   \[
   S_A + S_B + S_C \geq 3 \cdot \frac{k(k-1)}{2} = \frac{3k(k-1)}{2}
   \]

3. **Inequality for \( k \)**:
   Combining the two expressions, we get:
   \[
   kn \geq \frac{3k(k-1)}{2}
   \]
   Simplifying this inequality:
   \[
   2kn \geq 3k^2 - 3k
   \]
   \[
   2n \geq 3k - 3
   \]
   \[
   2n + 3 \geq 3k
   \]
   \[
   k \leq \frac{2n + 3}{3}
   \]
   Since \( k \) must be an integer, we take the floor function:
   \[
   k \leq \left\lfloor \frac{2n + 3}{3} \right\rfloor
   \]

4. **Verification for Specific \( n \)**:
   - If \( n = 3m \), then:
     \[
     k_{\text{max}} = 2m + 1
     \]
     Example sets:
     \[
     A = \{0, 1, \ldots, 2m\}, \quad B = \{2m, 2m-2, \ldots, 0, 2m-1, 2m-3, \ldots, 1\}, \quad C = \{m, m+1, \ldots, 2m, 0, 1, \ldots, m-1\}
     \]
   - For \( n = 3m + 1 \) and \( n = 3m + 2 \), similar constructions can be made based on the example above.

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