You could delay division by 2 until the end.
I.e ((2 * S mod N - n*(n+1) mod N) mod N) / 2
Where S is the sum of the array achieved through iterative addition with every addition mod N and N > 2n.
Where S is the sum of the array achieved through iterative addition with every addition mod N and N > 2n.
Edit: The simplest solution is probably just doing (n/2)*(n+1) (assuming n is even; move the division for n odd).