Let $n$ be an odd integer greater than $3$, and $\mathbb{S}=\{0, 1, \cdots, n-1\}$. Show that after removing any element from $\mathbb{S}$, it is always possible to equally divide the remaining elements in $\mathbb{S}$ into two groups such that their sum are congruent modulo $n$.
Let the element to be removed be $x\in\mathbb{S}$. Then transferring $\mathbb{S}\setminus\{x\}$ to set $\mathbb{T}$ by setting $t_i = (s_i - x)\pmod{n}$ will not change the sum of these element modulo $n$. This means that it is always possible to $\textit{shift}$ the elements without changing the nature of this problem. Therefore, it is sufficient to prove the claim with an assumption that the removed element is $0$.
When $n=4k+1$ where $k$ is a positive integer, then we can match the remaining $4k$ element into the following $2k$ pairs: $$\{1, 4k\}, \{2, 4k-1\}, \cdots, \{2k, 2k+1\}$$
The sum of every pair is $4k+1=n$, hence congruent $0$ modulo $n$. Therefore, selecting any $k$ pairs from them to form the first group and the remaining $k$ pairs to form the second group will satisfy the requirement because their sums will be both congruent to $0$ modulo $n$.
When $n=4k+3$, where $k$ is a positive integer, then we place $\{1, 2, 4k\}$ into the first group, and $\{3, 4k+1, 4k+2\}$ into the second group. Their sums are both congruent $0$ modulo $n=4k+3$. Next, matching the remaining elements into the following $(2k-2)$ pairs: $$\{4, 4k-1\}, \{5, 4k-2\}, \cdots, \{2k+1, 2k+2\}$$
The sum of every pair equals $(4k+3)$. Therefore, equally dividing these pairs into the two groups will satisfy the requirement.