Yeah, the parent comment was oversimplifying things. That particular solution only works for values of n != 3λ+1 (λ = 0,1,2,...). If you read the paper, they provide a number of different constructions, each one covering a different subset of n.

For n=4, I believe you'd need to use construction B, which is (k, 1+((2(k-1)+n-1) % 2n)) and (2n+1-k, 2n-((2(k-1)+n-1) % 2n)). Your resulting coordinates then become:

  (1,4) (2,6) (3,8) (4,2)
  (8,5) (7,3) (6,1) (5,7)

I think that should be a valid solution.