You actually spoiled a key trick that
you can group numbers into pairs of (1,6), (2,7), (3,8), (4,9), (5,10)
so I could easily come up with the answer of
1 6 2 7 3 | 8 4 9 5 10
10 5 9 4 8 | 3 7 2 6 1
---------------+---------------
2 7 3 8 4 | 9 5 10 1 6
6 1 10 5 9 | 4 8 3 7 2
---------------+---------------
3 8 4 9 5 | 10 1 6 2 7
7 2 6 1 10 | 5 9 4 8 3
---------------+---------------
4 9 5 10 1 | 6 2 7 3 8
8 3 7 2 6 | 1 10 5 9 4
---------------+---------------
5 10 1 6 2 | 7 3 8 4 9
9 4 8 3 7 | 2 6 1 10 5
which was generated in the following way:
First, place 1 6 2 7 3 8 4 9 5 10
on the first row. Then, on the 3, 5, 7, and 9th row, place the same sequence but starting at 2, 3, 4, 5 respectively. Note that rotating the row an even units does not break the condition, since the pairwise sums are still 7, 9, 11, 13, 15 in some order.
For each of the remaining rows, put the reverse of the row above. This trivially satisfies the Sudoku box requirement and pairwise sum requirement. The Sudoku column requirement is also met: you can see 1..5 and 6..10 are placed on even and odd rows (or vice versa) respectively.
As a bonus, there is an answer if you restrict the pair sums to only 11:
1 10 2 9 3 | 8 4 7 5 6
6 5 7 4 8 | 3 9 2 10 1
---------------+---------------
2 9 3 8 4 | 7 5 6 1 10
10 1 6 5 7 | 4 8 3 9 2
---------------+---------------
3 8 4 7 5 | 6 1 10 2 9
9 2 10 1 6 | 5 7 4 8 3
---------------+---------------
4 7 5 6 1 | 10 2 9 3 8
8 3 9 2 10 | 1 6 5 7 4
---------------+---------------
5 6 1 10 2 | 9 3 8 4 7
7 4 8 3 9 | 2 10 1 6 5