155045

The solution is n=8.
There are 4 even numbers in the sequence and 5 odd numbers. With 7 and 14 known there remains 3 even(e) numbers and 4 odd(o) ones.If the sum on column 3 is an even number (odd number on position 1 column 3) and therefor all other sums need to be even also, the careful consideration of all other empty spaces shows we reach an impossibility:
e o o
e o 7
e e 14 too many even numbers
or:
e o o
o e 7
o o 14 too many odd numbers
So the sum has to be an odd number with the first number in col 3 being even. That is we have either 8,10 or 12 in this position with 29,31 or 33 as the possible sums. Now looking at the last row (14 is in last position) we get the sum of the other 2 =15,17 or 19.Any 2 nr added are>16.So the first possibility is out. In the second case only 8+9=17, with 8 must be in lower left corner(even/odd considerations).This will force 13 to be in the middle which gives us 9 again on top of 2nd column=not possible.
We have proven sum is 33 with 12 upper left corner. Again looking at last row 2 numbers with the sum of 19. That can be 10+9 or 8+11. As before lower left corner has to be an even number(8 or 10).But 8 we already proven can't be in the lower left corner (see paragraph above). That is giving us in the lower left corner10, with 9 to the left and 11 in the middle. QED