Denote the vertices of the polygon as $V_1, \dots, V_n$ and let the real numbers in these vertices be $v_1, v_2, \dots, v_n$, respectively. Moreover, denote $V_{n+1} = V_1$ and $v_{n+1} = v_1$. Define the value of the side $V_iV_{i+1}$ of the polygon to be $s_i = v_{i+1} - v_i$.

Observe that $v_1, \dots, v_n$ satisfy the conditions of the problem if and only if the corresponding differences $s_1, \dots, s_n$ satisfy $\max\{|s_1|, \dots, |s_n|\} \le 1$ and $s_1 + \dots + s_n = 0$. Therefore the problem can be reformulated as finding the least non-negative real number $C$ such that, for arbitrary real numbers $s_1, \dots, s_n$, assumptions $\max\{|s_1|, \dots, |s_n|\} \le 1$ and $s_1 + \dots + s_n = 0$ would imply $\min\{|s_1|, \dots, |s_n|\} \le C$.

Clearly $\min\{|s_1|, \dots, |s_n|\} \le 1$ for any choice of $s_1, \dots, s_n$ satisfying the assumptions. If $n=2k$ for some $k \in \mathbb{N}$ then choosing $s_1 = \dots = s_k = 1$ and $s_{k+1} = \dots = s_{2k} = -1$ would imply $\min\{|s_1|, \dots, |s_n|\} = 1$. Hence $C=1$.

Let now be $n=2k+1$ for some $k \in \mathbb{N}$. Taking $s_1 = s_2 = \dots = s_{k+1} = \frac{k}{k+1}$ and $s_{k+2} = \dots = s_{2k+1} = -1$ establishes $\min\{|s_1|, \dots, |s_n|\} = \frac{k}{k+1}$. We show that $\min\{|s_1|, \dots, |s_n|\} \le \frac{k}{k+1}$ whenever $s_1, \dots, s_n$ satisfy the assumptions. To this end, suppose the contrary, i.e. $\min\{|s_1|, \dots, |s_n|\} > \frac{k}{k+1}$. Reorder $s_1, \dots, s_n$ as $d_1, \dots, d_n$ so that $d_1 \ge d_2 \ge \dots \ge d_n$. Note that there exists $l \in \{1, \dots, n\}$ such that $d_l > \frac{k}{k+1}$ and $d_{l+1} < -\frac{k}{k+1}$.

If $l \ge k+1$ then
$$0 = s_1 + \dots + s_n = d_1 + \dots + d_n \ge l \cdot d_l + (n-l)d_n > l \cdot \frac{k}{k+1} - (n-l) \ge (k+1) \cdot \frac{k}{k+1} - k = k - k = 0,$$
contradiction. In the case $l < k+1$, the proof is analogous (one can consider the opposite differences $-d_n, -d_{n-1}, \dots, -d_1$). Hence $C = \frac{k}{k+1} = \frac{\lfloor \frac{n}{2} \rfloor}{\lfloor \frac{n}{2} \rfloor + 1}$.