The numbers from one to nine taken three at a time formulate 63 combinations. From these numbers 1,3,5,7,9 are odd and the rest 2,4,6,8 are even. The even numbers taken two at a time must be paired in a particular way 2,6 and 4,8. If paired 2,8 and 4,6 then the number 5 has to be used two times to make the rows equal to 15. So let's complete the two rows 8,3,4 and 6,7,2. The remaining numbers formulate the third row 1,5,9. Now we have the 3 rows.
8 3 4
1 5 9
6 7 2
All the columns and the rows sum to 15.
For unlimited solutions for the 3x3 grid which follows the rules set in the question you apply the method I used in my solution. If you multiply the number 9 by any natural number greater than zero then you obtain the next 3x3 grid.Let's have nx9: if n=2 then we have $\frac{(18X9)-(9X10)}{6}=42$ which is the sum for rows, columns, diagonals. We have 10,11,12,13,14,15,16,17,18. Now we have 5 even and 4 odd numbers. There is only one way to pair the odd numbers 13,17 and 11,15. Inserting the even numbers 12 and 16 we have 13,12,17 and 11,16,15. The remaining even numbers make the third row.
13 12 17
18 14 10
11 16 15
all add to 42.