Let the vertices of an $n$-gon $A_0A_1\dots A_{n-1}$ be labeled with numbers satisfying the conditions. Let $a$ be the least among these numbers. W.l.o.g., assume that $A_0$ contains $a$ and the indices of vertices are increasing counterclockwise. Let $b$ and $c$ be the numbers at vertices $A_{n-1}$ and $A_{n-2}$, respectively (Fig. 8). By condition (1), $a = |b-c| \ge 0$. By the choice of $a$, we must have $b \ge a$, whence by condition (1), $A_1$ contains $b-a$. As $b-a \ge a$ by the choice of $a$, the condition (1) also implies that $A_2$ contains $b-2a$. Hence $A_3$ contains $(b-a) - (b-2a) = a$ by condition (1) and non-negativity of $a$. Since the vertices can be renumerated without changing the direction in such a way that $A_3$ becomes $A_0$, we can conclude that $A_6$ also contains $a$. Similarly, every third vertex contains $a$.

If the number $n$ of vertices is not divisible by 3 then either $A_{n-1}$ or $A_1$ must contain $a$ and, as we can repeat this argument, also $A_{n-2}$ or $A_2$, respectively, contains $a$. Thus three consecutive vertices contain $a$. Applying the condition (1) to these three vertices, we obtain $a = |a-a| = 0$. But 0 being in two consecutive vertices implies that 0 is in all vertices. Then the sum of all labels is 0, contradicting the condition (2).

This shows that $n$ must be divisible by 3. Let $n = 3k$ where $k$ is a positive integer. For every $3k$-gon, the conditions of the problem can be satisfied by writing 0 into every third vertex and $\frac{1}{2k}$ into all other vertices (Fig. 9 depicts the situation for $k=4$).