Different sums in a $3\times 3$ grid 
We write the numbers $1, 2, 3$ in the cells of grid $3 \times 3$ and we calculate the sums in rows and columns. What is the maximal number of different sums?

I am learning combinatorics, and this is a problem I got stuck with. I cannot show my work, since I am new here and am not fluent in mathjax.
 A: The possible sums range from $3$ to $9$. This gives $4$ odd numbers and $3$ even numbers in range.
Consider two cases - 1) we have both $3$ and $9$ as sums, using $(1,1,1)$ and $(3,3,3)$. Then these are either both on rows or both on columns, since they cannot intersect. Say they are two rows; then the third row can be all different ($1,2,3$), in which cases we have two totals of $6$, or not all different, in which we have a matching pair of column totals. In either case the maximum number of different values is $5$.
Case 2) - we don't have both $3$ and $9$. Now there are $3$ odd totals and $3$ even totals to aim for, but we cannot have all of them, since then the total of all rows and columns would be odd, which is not possible since that is twice the total grid sum and must be even. So again we cannot have $6$ different row and column sums, and it's easy to demonstrate grids with $5$ different sums.
So the maximal number of different row and column sums from such a grid is $5$.
A: For each row/column the minimum sum which is reachable is [1,1,1] => 3 
and going the same way the maximum sum which can be reached is [3,3,3] => 9 
So All sums of range [3,9] are achievable.. I hope you can find the answer now ..
