Number Theory: Prove that there are two different permutations $a, b$ which $n! \ | \ S(a)-S(b)$ 
Assume $n$ is an integer which is odd and $n > 1$. We have function
  $S(a)$ which $a$ is a permutation of numbers $\{1,2 \ldots ,n\}$  and $S(a) =
> c_1a_1 + c_2a_2 + ... + c_na_n$ ($c_i$'s are all constants)
Prove that there are two different permutations $a, b$ which:
$n! \ | \ S(a)-S(b)$

How can i prove this statement?
 A: Assume $c_1,...,c_n$ are integers, and $n>1$ is an odd integer.

Let $P$ be the set of all permutations of $\{1,...n\}$.

The claim is that there exist distinct $a,b \in P\,$ such that $n!\mid (S(a)-S(b))$.

Suppose instead that no such pair $a,b$ exists. 

Our goal is to derive a contradiction.

Let $x = {\displaystyle{\sum_{p \in P}S(p)}}$. Then by symmetry,
\begin{align*}
x &= (c_1 + c_2 + \cdots + c_n)\,\bigl((n-1)!\bigr)\,(1 + 2 + \cdots + n)\\[4pt]
&= (c_1 + c_2 + \cdots + c_n)\,\bigl((n-1)!\bigr)\!
\left(
{\small{\frac{n(n+1)}{2}}}
\right)\\[4pt]
&= (c_1 + c_2 + \cdots + c_n)\,(n!)\left({\small{\frac{n+1}{2}}}\right)\\[4pt]
\end{align*}
hence $x \equiv 0 \pmod{n!}$.

But by assumption, if $a,b \in P$ with $a \ne b$, then $S(a) \not\equiv S(b) \pmod{n!}$,
hence 
$$\{S(p)\;\text{mod}\;n!\} = \{0,1,2,...,n!-1\}$$

Let $w = n!/2.\;\,$Then

\begin{align*}
&x \equiv 0 \pmod{n!}\\[4pt]
\implies\; &0 + 1 + 2  + \cdots + (n!-1) \equiv 0 \pmod{n!}\\[4pt]
\implies\;  &0 + (1 + (n! - 1)) + (2 + (n! - 2)) + \cdots + ((w-1) + (w+1)) + w  \equiv 0 \pmod{n!}\\[4pt]
\implies\; &w  \equiv 0 \pmod{n!}\\[4pt]
\end{align*}
contradiction, since $0 < w < n!$.

The contradiction proves the claim.
