For each $i \in \{1, 2, \dots, 2021\}$, let $e_i = (\underbrace{1, \dots, 1}_{i-1}, \underbrace{-1, \dots, -1}_{2021-i})$.
For two sequences $a = (a_1, a_2, \dots, a_{2020})$, $b = (b_1, b_2, \dots, b_{2020})$, we define
$$a \cdot b = a_1 b_1 + a_2 b_2 + \dots + a_{2020} b_{2020}.$$
It's easy to verify that $a \cdot b$ is even for each $a, b \in F$. Let $a = (a_1, a_2, \dots, a_{2020}) \in F$ and denote $b_i = a \cdot e_i$ for each $i \in \{1, 2, \dots, 2020\}$, then
$$b_{i+1} - b_i = a \cdot e_{i+1} - a \cdot e_i = 2a_i \Rightarrow |b_{i+1} - b_i| = 2, \forall i \in \{1, 2, \dots, 2020\}.$$
Let $S = \{e_1, e_2, \dots, e_{2020}\}$. It is clear that $b_{2021} = -b_1$ and $b_i$ is even for each $i \in \{1, 2, \dots, 2020\}$. If $b_1 \neq 0$, there exists an integer $k$ such that $1 < k < 2021$ and $b_k = 0$. Hence, $S$ satisfies the problem's requirements. $\square$