Answer: all even $n$.
The following example shows that $n = 2k + 1$ does not satisfy the problem condition. We number clockwise the vertices of the $n$-gon as $A_0, A_1, \dots, A_k, A_{-k}, \dots, A_{-1}$ and place the number $|i|$ at the vertex $A_i$. Since $n$ is odd, the pairs of antipodal points have the form «vertex-midpoint of an arc». But the integer numbers are placed at the vertices while the fractional numbers are placed at the midpoints of all arcs (except for the arc $A_k A_{-k}$ at the midpoint of which the number $k \neq 0$ is placed).
Show that for $n = 2k$ there exist antipodal points with the same numbers. We number clockwise the vertices of the $n$-gon as $A_1, A_2, \dots, A_{2k}$, and let $a_1, a_2, \dots, a_{2k}$ be the numbers at these vertices, respectively. For convenience we assume that $A_{2k+1} = A_1$ and $a_{2k+1} = a_1$. Consider the numbers
$$a_{k+1} - a_1, a_{k+2} - a_2, \dots, a_{k+i} - a_i, \dots, a_{k+(k+1)} - a_{k+1}. \quad (1)$$
The absolute value of the difference between any two adjacent terms of sequence (1) is less than or equal to 2. Indeed,
$$|(a_{k+i} - a_i) - (a_{k+i+1} - a_{i+1})| \le |a_{k+i+1} - a_{k+i}| + |a_{i+1} - a_i| \le 1+1=2, \quad (2)$$
since, by condition, $|a_{k+i+1} - a_{k+i}| \le 1$, $|a_{i+1} - a_i| \le 1$ for all $i$ and $k$. If there is zero between the terms of (1), then there are antipodal points with the same numbers.
Suppose that there is no zero in sequence (1). Since the first and the last terms of (1) have different signs (their sum is equal to 0), there are two adjacent terms $a_{k+i} - a_i$ and $a_{k+i+1} - a_{i+1}$ with different signs. Since all terms of (1) are integer, 0 does not belongs to (1), and (2) holds, we see that exactly one of these two adjacent numbers is equal to 1 while the other is equal to -1. Therefore, their sum $a_{k+i} - a_i + a_{k+i+1} - a_{i+1} = 1 + (-1) = 0$, i.e., $a_i + a_{i+1} = a_{k+i} + a_{k+i+1}$. Hence the numbers placed at the antipodal points (midpoints of the arcs $A_{k+i}A_{k+i+1}$ and $A_iA_{i+1}$) are equal. Thus all even $n$ satisfy the problem condition.