A divisible subset

I got the following from a book:

Let $n \in \mathbb{N}$ . Show that for every set S of $n$ integers, there is a nonempty subset $\color{blue}T$ of $\color{blue}S$ such that $\color{blue}n$ divides the sum of elements of $\color{blue}T$ .

$\text{Proof:}$

Let $S_k=\{a_1,a_2,...,a_k\}$ for each integer $k$ with $1\leq k \leq n$ . For each integer $k(1\leq k \leq n),$

$\Sigma_{i=1}^k a_i\equiv r \: \text{(mod n)}$ for some integer $r$ , where $0 \leq r \leq n-1$ .

We consider two cases:

Case 1:

$\Sigma_{i=1}^k a_i \equiv 0\: \text{(mod n)}$ for some integer $k$ . Then $n|\Sigma_{i=1}^k$ , that is, $n$ divides the sum of the elements $S_k$ .

Case 2:

$\Sigma_{i=1}^k a_i \not\equiv 0\: \text{(mod n)}$ for all integer $k (1\leq k \leq n)$ .

Hence there exist integers there exist integers $s$ and $t$ with $1<s<t\leq n$ such that $\Sigma_{i=1}^s \equiv r\: \text{(mod n)}$ and $\Sigma_{i=1}^t \equiv r\: \text{(mod n)}$ for integer $r$ with $1\leq r \leq n-1$ .

Therefore,

$\Sigma_{i=1}^s a_i \equiv \Sigma_{i=1}^t a_i \: \text{(mod n)}$

and so

$n|(\Sigma_{i=1}^t a_i - \Sigma_{i=1}^s a_i )$

Hence

$n|\Sigma_{i=s+1}^s a_i )$

Remark:

1. I have tested a set of $n$ elements on the computer and found that at least one of the $2^n-1$ nonempty subsets is divisible by $n$ .

2. In regard to case 2 in the proof above, I can't see how

$\Sigma_{i=1}^k a_i \not\equiv 0\: \text{(mod n)}$ implies the existence of $\Sigma_{i=1}^s \equiv r\: \text{(mod n)}$ and $\Sigma_{i=1}^t \equiv r\: \text{(mod n)}$ . It might be true by chance.

Question: Does anyone here have a better proof than this?