For what values of $ n$ does there exist an $ n \times n$ array of entries -1, 0 or 1 such that the $ 2 \cdot n$ sums obtained by summing the elements of the rows and the columns are all different?
1. **Construction for \( n = 6 \):**
   We start by constructing a \( 6 \times 6 \) array with entries -1, 0, or 1 such that the sums of the rows and columns are all different. Here is an example of such an array:
   \[
   \begin{bmatrix}
   \textcolor{red}{1} & -1 & -1 & -1 & -1 & -1 \\
   1 & \textcolor{red}{0} & -1 & -1 & -1 & -1 \\
   1 & 1 & \textcolor{red}{1} & -1 & -1 & -1 \\
   1 & 1 & 1 & \textcolor{red}{0} & -1 & -1 \\
   1 & 1 & 1 & 1 & \textcolor{red}{1} & -1 \\
   1 & 1 & 1 & 1 & 1 & \textcolor{red}{0}
   \end{bmatrix}
   \]
   It is not hard to verify that the sums of the rows and columns are all different.

2. **Proof for \( n \) odd:**
   We need to show that for odd \( n \), it is impossible to construct such an array. Assume \( n \) is odd and consider an \( n \times n \) array with entries -1, 0, or 1. We aim to show that the sums of the rows and columns cannot all be different.

3. **Assumptions and Setup:**
   Without loss of generality, we can assume that the sum of the elements in the first row is non-negative. If necessary, we can multiply every element of the matrix by -1 to ensure this. Let \( m \) be the sum of the first row, so \( m \geq 0 \). Note that \( m \in [-n, n] \).

4. **Swapping Rows and Columns:**
   We can swap rows and columns to simplify our analysis. Assume that the \( k \) leftmost columns have non-negative sums, and the \( n-k \) rightmost columns have negative sums. Similarly, assume that the \( n-k \) top rows have non-negative sums, and the \( k \) bottom rows have negative sums.

5. **Defining Regions:**
   Split the grid into four regions:
   - A top-left \( (n-k) \times k \) region \( A \) with sum \( a \)
   - A top-right \( (n-k) \times (n-k) \) region \( B \) with sum \( b \)
   - A bottom-right \( k \times k \) region \( C \) with sum \( c \)
   - A bottom-left \( k \times (n-k) \) region \( D \) with sum \( d \)

6. **Double-Counting Sums:**
   By double-counting the sums over regions and rows, we have:
   \[
   2a + b + d = \frac{n(n+1)}{2} - m
   \]
   and
   \[
   2c + b + d = -\frac{n(n+1)}{2}
   \]
   Subtracting these two equations, we get:
   \[
   2a - 2c = n^2 + n - m
   \]

7. **Bounding the Left-Hand Side:**
   The left-hand side \( 2a - 2c \) is bounded above by \( 4k(n-k) \) because the total area covered by regions \( A \) and \( C \) is \( 2k(n-k) \). Since \( n \) is odd, this quantity is at most:
   \[
   4 \cdot \frac{n-1}{2} \cdot \frac{n+1}{2} = n^2 - 1
   \]

8. **Contradiction:**
   However, the right-hand side \( n^2 + n - m \geq n^2 \), which is a contradiction because \( n^2 \) is greater than \( n^2 - 1 \). Therefore, it is impossible to construct such an array for odd \( n \).

\(\blacksquare\)

The final answer is \( \boxed{ n } \) even only.