The main idea is to use complex numbers. Let $x$ be a primitive $n$th root of unity. Normally I would use Greek letters for this, but I don't feel like it. Erase everything that is outside the rectangle. At each point $(i,j)$ which is either INSIDE or ON THE BOUNDARY of the rectangle, we erase the existing label (which is $i+j \pmod{n}$) and write a new label (which is $x^{i+j}$). The key property is that the sum of the remaining labels must be 0 (hint: combine properties (i), (ii)). But we can also write the sum of the remaining labels as $(x^0 + x^1 + x^2 + \cdots + x^a)(x^0 + x^1 + x^2 + \cdots + x^b)$; hence at least one of $x^0 + x^1 + x^2 + \cdots + x^a,x^0 + x^1 + x^2 + \cdots + x^b$ must be equal to 0. This implies $b \equiv -1 \pmod{n}$ or $a \equiv -1 \pmod{n}$. Conversely, it's obvious that both of these cases work. By property (i), we must have that $n|a+b$, so we are done.

Say an $A \times B$ grid of $AB$ lattice points is good if each remainder occurs an equal number of times among the labels. Note that goodness is invariant under shifting. I claim the rectangle is good iff $n \mid A$ or $n \mid B$. The if direction is clear. For only if, suppose we shift a good $A \times B$ rectangle one unit to the left; then the uncovered column of $B$ labels must contain the same labels as the newly covered column, i.e. $\{x+1, x+2, \dots, x+B\} \to \{(x-A)+1, (x-A)+2, \dots, (x-A) + B\}$ are the same modulo $n$. But this implies either $n \mid A$ or $n \mid B$, as desired. (imagine rotating an arc around a circle.) So, for the problem at hand, the necessary and sufficient condition is ($n \mid a-1$ or $n \mid b-1$) AND ($n \mid a+1$ or $n \mid b+1$).