Please post your solutions. This is just a solution template to write up your solutions in a nice way and formatted in LaTeX. But maybe your solution is so well written that this is not required finally. For more information and instructions regarding the ISL/ILL problems please look here: introduction for the IMO ShortList/LongList project and regardingsolutions

Ah...at last I solved an IMO-Combinatorics problem. We consider all $n!$ Permutations $a$ of ${1,\ldots,n}$ and the corresponding $S(a)$ modulo $n!$. If two of them are congruent $\mod{ n!}$, then we are done. So let's suppose that all $n!$ permutations are in different residues. Sum up over $S(a)$ and we have \[ \sum_{a}S(a)\equiv 0+1+\ldots+n!-1 \equiv \frac{(n!-1)n!}2\mod {n!} \] But on the other hand, we have \[ \sum_{a}S(a)=\sum_{i=1}^nc_i( (n-1)!\cdot (1+2+\ldots+n)) = \sum_{i=1}^nc_i( n! \cdot \frac{n+1}2) \] Since $n$ is odd, we have $2\mid n+1$, thus $\sum_a S(a)\equiv 0 \mod{n!}$, so we have \[ \frac{(n!-1)n!}2\equiv 0\mod{n!} \] which is a contradiction since $n!$ is even.

Suppose for the sake of contradiction that for all permutations $a$ and $b$ of $\{1,2, \ldots, n\}$, we have that $S(a) \neq S(b) \pmod{n!}$. There are $n!$ permutations of $\{1,2, \ldots, n\}$ and $n!$ residue classes modulo $n!$, so for every integer $m$ with $0 \leq m < n!$, there is a unique permutation $a$ of $\{1,2,\ldots,n\}$ such that $S(a) \equiv m \pmod{n!}$. Let $k = \frac{(n+1)(c_1 + c_2 + \cdots + c_n)}{2}$. We can find some permutation $a = (a_1, a_2, \ldots, a_n)$ such that $S(a) \equiv c_1 a_1 + c_2 a_2 + \cdots + c_n a_n \equiv k \pmod{n!}$. But the permutation $a' = (n+1-a_1, n+1-a_2, \ldots, n+1-a_n)$ also satisfies $S(a) \equiv (n+1)(c_1 + c_2 + \cdots + c_n) - (c_1 a_1 + c_2 a_2 + \cdots + c_n a_n) \equiv 2k - k \equiv k \pmod{n!}$. Hence, $a'$ and $a$ must be the same permutation, which is absurd.

Let $\mathcal{P}_1, \mathcal{P}_2, \cdots , \mathcal{P}_{n!}$ denote the $n!$ permutations of $(1, 2, \cdots , n).$ We shall take the indices modulo $n!,$ meaning that $\mathcal{P}_x = \mathcal{P}_{x \pmod{n!}}.$ Also, let $\mathcal{D}_{n,x}$ denote the $x$th digit of $\mathcal{P}_n.$ Suppose the problem statement isn't true. Then, \[S(\mathcal{P}_x) \equiv S(\mathcal{P}_y) \pmod{n!} \iff x \equiv y \pmod{n!},\] implying that $\mathcal{P}_i$ is bijective modulo $n!,$ or in other words, $(S(\mathcal{P}_1), S(\mathcal{P}_2), \cdots , S(\mathcal{P}_{n!}))$ is equivalent to $(1, 2, \cdots , n!)$ modulo $n!$ in some order. We then have \[\displaystyle \sum_{i=1}^{n!} S(\mathcal{P}_i) \equiv \displaystyle \sum_{i=1}^{n!} i \equiv \dfrac{n!(n!+1)}{2} \pmod{n!}.\] But we also have \begin{align*} \displaystyle \sum_{i=1}^{n} S(\mathcal{P}_i) & \equiv \displaystyle \sum_{i=1}^{n} \left( \displaystyle \sum_{j=1}^{n} c_j \mathcal{D}_{i,j} \right) \\ & \equiv (n-1)! \displaystyle \sum_{j=1}^{n} c_j \left( \displaystyle \sum_{k=1}^{n} \mathcal{D}_{j,k} \right) \\ & \equiv \dfrac{n(n+1)}{2} (n-1)! \cdot \left( \displaystyle \sum_{j=1}^{n} c_i \right) \\ & \equiv 0 \pmod{n!},\end{align*} implying $\dfrac{n!(n!+1)}{2} \equiv 0 \pmod{n!} \implies 2 \mid n!+1,$ contradiction. $\blacksquare$

Wow, this problem is really nice. Unfortunately my solution seems to be pretty similar compared to previous ones in the thread FTSOC $S(1), S(2), \cdots, S(n!)$ are all distinct residues $\pmod{n!}$. Let's consider $S=\sum_{j=1}^{n!} \sum_{i=1}^{n} c_ia_i$ in two different ways. First, summing over $S(1), \cdots, S(n!)$ we have $S=1!+2!+\cdots+(n!-1) \equiv \frac{n!(n!-1)}{2} \pmod{n!}$, which isn't $0 \pmod{n!}$. Considering $S$ a different way, observe that there are $(n-1)!$ times where $c_i$ has coefficient $1, 2, \cdots, n$, $1 \le i \le n$. Summing up over the $c$'s, we have $S=(n-1)! \frac{n(n+1)}{2} \sum_{i=1}^{n} c_i \equiv 0 \pmod{n!}$. This is our contradiction and we are done.

I apologize to everyone for my solution being similar to above solution. Let $S_{n}$ be set of all permutation of $(1,\dots ,n)$.We suppose that there is no $a\neq b$ which satisfy $S(a)\equiv S(b)\mod{n!}$.Then $\{S(a)\}_{a\in S_{n}}{}$ is $\{0,1,\dots ,n!-1\}$ in$\mod{n!}$.We calculate $\sum_{a\in S_{n}}{}S(a)$ in $2$ ways. First $\sum_{a\in S_{n}}{}S(a)\equiv \sum_{k=0}^{n!-1}k\equiv \frac{(n!-1)(n!)}{2}\mod{n!}$. Second The number of permutations which $a_i=k(1\le i,k\le n)$ is $(n-1)!$.Thus $\sum_{a\in S_{n}}{}S(a)\equiv \sum_{i=1}^{n}c_{i}\cdot (n-1)!\cdot (1+\dots +n)$ $\equiv n!\frac{n+1}{2}\sum_{i=1}^{n}c_{i}\equiv 0\mod{n!}$. From First and Second, $\frac{(n!-1)(n!)}{2}\equiv 0\mod{n!}\Leftrightarrow \frac{n!}{2}\equiv 0\mod{n!}$ which is absurd.Therefore ∃$a\neq b$ s.t. $n! | S(a)-S(b)$.$\blacksquare$

ISL 2001 C2 wrote: Let $n$ be an odd integer greater than 1 and let $c_1, c_2, \ldots, c_n$ be integers. For each permutation $a = (a_1, a_2, \ldots, a_n)$ of $\{1,2,\ldots,n\}$, define $S(a) = \sum_{i=1}^n c_i a_i$. Prove that there exist permutations $a \neq b$ of $\{1,2,\ldots,n\}$ such that $n!$ is a divisor of $S(a)-S(b)$. Solution. Let $\mathcal{P}$ be the set of all permutations of $\{1,2,3,\ldots,n\}.$ Assume for the sake of contradiction that there doesn't exist such $a\ne b.$ Consider the multiset $T=\{S(\pi)\mid \pi\in\mathcal P\},$ because of our assumption and $|T|=n!$ , when we take $T$ modulo $n!$ we get the complete set of residues modulo $n!~.$ Let $X=(x_1,x_2,x_3,\ldots,x_n)$ be the permutation such that \[S(X)\equiv \left(\frac{n+1}{2}\right)\left(\sum_{i=1}^{n}c_i\right)\pmod{n!}\]Define $\overline{X}=(n+1-x_1,n+1-x_2,\ldots,n+1-x_n),$ note that $X\ne \overline{X}$ and that $\overline{X}$ is a valid permutation. Now, \[S(\overline{X})=\sum_{i=1}^n c_i(n+1-x_i)=(n+1)\sum_{i=1}^n c_i-S(X)\equiv (n+1)\sum_{i=1}^n c_i- \left(\frac{n+1}{2}\right)\left(\sum_{i=1}^{n}c_i\right)\pmod{n!}\equiv S(X)\pmod{n!}\]Which is a contradiction to our assumption. And we are done. $\blacksquare$

Suppose for sake of contradiction that this didn't happen. Then, all the $S(\sigma)\pmod{n!}$ are different for all $\sigma\in S_n$. Choose $\sigma$ uniformly at random. By linearity of expectation, we have \[\mathbb{E} S(\sigma)=c_i\frac{n+1}{2},\]so \[\sum_{\sigma\in S_n}S(\sigma)=\frac{(n+1)!}{2}\sum_{i=1}^n c_i\equiv 0\pmod{n!}\]since $\frac{n+1}{2}\in\mathbb{Z}$. By the hypothesis, we have \[\sum_{\sigma\in S_n}S(\sigma)\equiv\sum_{k=0}^{n!-1}k\equiv\frac{n!(n!-1)}{2}\pmod{n!},\]which isn't divisible by $n!$ since $n!-1$ is odd (this is why we need $n>1$), This is a contradiction, as desired. $\blacksquare$

Assume FTSOC that all the $S(\pi)$'s are distinct mod $n!$ for all $\pi \in S_n$. Then \[ \sum_{\pi \in S_n} S(\pi) \equiv 0+\cdots+(n!-1) = \frac{(n!-1)(n!)}{2} \not \equiv 0 \pmod{n!} \]since $n!$ is even for $n>1$. Treat the permutations and $\{c_i\}$ as vectors in $\mathbb{F}_{n!}^n$. Note that the vector sum of all the permutations is \[ ((n-1)!\cdot (1+\cdots+n))(1,\ldots,1) = \frac{n!(n+1)}{2}(1,\ldots,1).\]Then \begin{align*} \sum_{\pi \in S_n} S(\pi) &= \sum_{\pi \in S_n} \vec{c}\cdot \pi \\ &= \vec{c} \cdot \sum_{\pi \in S_n} \pi \\ &= \vec{c} \cdot \frac{n!(n+1)}{2} (1,\ldots,1) \\ &= \frac{n!(n+1)}{2} \left(\sum c_i \right) \equiv 0 \pmod{n!} \end{align*}since $n$ is odd. Contradiction!

For storage. Suppose not. Then let $\mathcal{P}$ be the set of all permutations $a$. Now for all permutations, their residues are distinct modulo $n!$. Hence \begin{align*} 0 + 1 + \dots + (n! - 1) &\equiv \sum_{a\in\mathcal{P}} S(a) \\ &\equiv \frac{n(n+1)}{2}\cdot(n-1)!\cdot\sum_{i=1}^n c_i \end{align*}by looking at the amount of times $1$, $2$, $\dots$, $n$ appears in the $i$th slot. But \[\frac{n(n+1)}{2}\cdot(n-1)!\cdot\sum_{i=1}^n c_i = \frac{n+1}{2}\cdot n!\cdot \sum_{i=1}^n c_i\]and $n$ is odd, so the RHS is congruent to $0$ modulo $n!$. Hence we have $\frac{(n!-1)(n!)}{2} \equiv 0 \mod{n!}$. But $n!$ is even, so this is equivalent to $-\frac{n!}{2}$, which is clearly nonzero modulo $n!$, contradiction.

For the sake of contradiction assume this is not the case, which implies that $S(a)$ takes on every residue modulo $n!$ as it ranges over all permutations. Letting $P$ be the set of all permutations, we have that \begin{align*} \sum_{k \in P} S(k) &\equiv 0+1+2+\cdots + (n!-1)\\ &\equiv \frac{(n!-1)(n!)}{2} \pmod{n!}. \end{align*} If we compute \begin{align*} \sum_k S(k) &= \sum^n_{j=1} c_j \times (n-1)! \times (1+2+3+\cdots+n)\\ &= \sum_{j=1}^n c_j\left[(n!)\left(\frac{n+1}{2}\right)\right]. \end{align*}So, $\sum_k S(k) \equiv 0 \mod{n!}.$ Thus we can deduce that $$\frac{(n!-1)(n!)}{2} \equiv 0 \pmod{n!},$$which is a contradiction since $n!$ is even. $\blacksquare$

AFTSOC that all $S(a)\neq S(b)$, then the set of $S(p)$ is the residues from $0,1,\ldots, n!-1$. Let $P$ be the set of all permutations, then \[\sum_{p\in P} S(p) \equiv \frac{(n!-1)(n!)}{2} \not\equiv 0 \pmod{n!}\]however, \[\sum_{p\in P} S(p) = n!\cdot \frac{1+2+\cdots n}{n}\cdot \sum c_i = n!\cdot \frac{n+1}{2}\sum c_i \equiv 0 \pmod{n!}\]a contradiction, so we are done. $\blacksquare$

Probably similar to the above solutions, posting just for storage

Assume for the sake of contradiction that $n!$ does not divide $S(a)-S(b)$, this means all the $S(i)$ give distinct remainder upon division by $n!$. $$\implies \sum_{\text{over all permutations}} S(a) \equiv 0 + 1 + \cdots + n!-1 \equiv \dfrac{n!(n! - 1)}{2} \equiv \frac{n!}{2} \pmod{n!}$$On the other hand $$\sum_{\text{over all permutations}} S(a) = \sum_{i=1}^n c_i \cdot (n-1)! \cdot (1 + 2 + \cdots n) = \dfrac{n!(n+1)}{2} \sum_{i=1}^n c_i \equiv 0 \pmod{n!}$$because $\dfrac{n+1}{2} \in \mathbb Z$. So $\dfrac{n!}{2} \equiv 0 \pmod{n!}$ which is a contradiction.

The statement clearly is not true for $n = 1$ or $n = 2$, but one may check (with a computer) that it does hold for $n = 4$. I wonder whether the result is true for all even $n \ge 4$. Of course, the above proofs fail in this case, so one needs something more sophisticated.

Let $P$ denote the set of permutations of $\{1,2,\ldots,n\}$. Since $|P|=n!$, we only have to show that we cannot have $\{S(\sigma)\}_{\sigma \in P}\equiv \{1,\ldots,n!\} \pmod{n!}$. Suppose otherwise. Then summing over $1,\ldots,n!$ we find that $$\sum_{\sigma \in P} S(\sigma)\equiv \frac{n!(n!+1)}{2}\equiv \frac{n!}{2} \pmod{n!},$$but summing over $P$ directly, we obtain $$\sum_{\sigma \in P} S(\sigma)=\sum_{\sigma \in P}\sum_{i=1}^n c_i\sigma(i)=\sum_{i=1}^n c_i\left(\sum_{\sigma \in P}\sigma(i)\right)=\frac{n(n+1)(n-1)!}{2}\sum_{i=1}^nc_i\equiv 0 \pmod{n!},$$as $n$ is odd. Since $\tfrac{n!}{2} \not \equiv 0 \pmod{n!}$, we have a contradiction. $\blacksquare$

Suppose by way of contradiction that there exist no two permutations $a$ and $b$ such that $S(a) \equiv S(b) \mod{n!}$ This means that all permutations have a unique residue modulo $n!$ Thus, if we compute the sum of $S(p)$ modulo $n!$ where $p$ ranges over all permutations of $[1, 2, 3, 4, \dots n]$, we will obtain $\sum_{i = 0}^{n!-1}i \equiv \frac{(n!-1)n!}{2} \equiv \frac{-n!}{2} \equiv \frac{n!}{2} \pmod{n!}$ Now let us compute this exact sum modulo $n!$ in another way. We can think about how each of $c_1, c_2, c_3 \dots c_n$ contribute to the sum. Observe that each of the $c_i's$ have a "contribution" of $(n-1)!(1+2+3+4+\dots+n) = (n-1)!\frac{n(n+1)}{2}c_i$. So the sum we desire modulo $n!$ is $\sum_{i = 1}^{n}((n-1)!\frac{n(n+1)}{2}c_i) \equiv (n-1)!\frac{n(n+1)}{2}\sum_{i=1}^{n}c_i \equiv 0 \equiv \frac{n+1}{2}n!\sum_{i=1}^{n}c_i \pmod{n!}$. Since $n$ is odd, we know that $\frac{n+1}{2} = k$ where $k$ is some integer. So, we have effectively deduced that the sum of $S(p)$ where $p$ iterates over the permutations of $[1, 2, 3, \dots n]$ is $0 \pmod{n!}$. But note that we claimed earlier that the sum was actually $\frac{n!}{2} \pmod{n!}$, therefore creating a contradiction.

Seriously, this was an IMO problem? Assume for the sake of contradiction that all values of $S(a)$ are distinct mod $n!.$ The main idea is to add everything together. On one hand, we are adding all of the distinct residues mod $n!,$ which gives $1 + 2 + \dots + n! = \frac{n! (n! + 1)}{2}.$ Noting that $n! + 1$ is odd and $n!$ is even, this is equivalent to $\frac{n!}{2} \pmod{n!}.$ On the other hand, for each $c_i,$ each possible value of $a_i$ from $1$ to $n$ inclusive is covered $(n-1)!$ times, so the sum is equivalent to $$(c_1 + c_2 + \dots + c_n)(n-1)!(1+2+\dots + n) = (c_1 + c_2 + \dots + c_n)(n-1)! \cdot \frac{n(n+1)}{2} = \frac{(c_1 + c_2 + \dots + c_n)(n+1)!}{2}.$$Note that $\frac{(n+1)!}{2} = \frac{n!}{2} \cdot (n+1).$ Since $n$ is odd, $n + 1$ is even, which makes this equivalent to $0 \pmod{n!}.$ Therefore, the entire thing is equivalent to $0 \pmod{n!},$ which is a clear contradiction. Therefore, by Pigeonhole, at least one residue class is covered by $S$ at least twice.

Otherwise $S(a)$ are distinct and the sum of $S(a)$ over all $a$ will be $\equiv \frac{n!}2\pmod{n!}.$ However it is not hard to show that this sum is $n!\left(\frac{n+1}2\right)\sum_{i=1}^n c_i\equiv 0\pmod{n!}.$

for this to be false the values of $S(a)$ have to be from $0$ to $n!-1$ mod $n!$ but then when you sum it you get that it is not $0$ mod $n!$ from the mods, but from the $c$s, it is $0$ mod $n!$, so there is a contradition so that means that this statement is true

Let us put the permutations as $\sigma_1,\ldots,\sigma_{n!}$. Then if no two $S(\sigma_i) \equiv S(\sigma_j) \mod{n!}$, we have that $S(\sigma_1),\ldots,S(\sigma_{n!})$ form a complete residue system of $n!$, that means $$\sum_{i=1}^{n!} S(\sigma_i) \equiv \frac{n!(n!-1)}{2} \mod {n!}$$but on the other hand we could also evaluate $$\sum_{i=1}^{n!} S(\sigma_i) \equiv \frac{n!(n+1)}{2}(c_1+\ldots+c_n) \equiv 0 \mod{n!}$$as $n+1$ is even. Thus since $\gcd(n!-1,n!)=1$, we must have $\frac{n!}{2} \equiv 0 \mod{n!}$ which is absurd as $3 \leq \frac{n!}{2}<n!$.

Suppose FTSOC there does not exist two permutations $a$, and $b$ such that $S(a)- S(b) \equiv 0 \pmod{n!}$. Observe that $\sum_{a \in \text{perms}} S(a) = \frac{n!}{2}(n+1)(c_1+c_2 + \cdots + c_n)$ However as $n+1$ is even this is divisible by $n!$. Now notice that since all $n!$ different $S(a)$'s have a distinct remainder modulo $n!$ then $\sum_{a \in \text{perms}} S(a) \equiv \frac{n!(n!-1)}{2} \equiv -\frac{n!}{2} \pmod{n!}$ which is not divisible by $n!$, which contradicts the divisibility from earlier meaning we can conclude.

It suffices that $S$ over all $n!$ permutations taking on values $0$, $1$, $\dots$, $n!-1$ in $\pmod{n!}$ is impossible. Let $C$ be the sum of all $c_i$. The sum of $S(a)$ over all permutations $a$ is then equal to $C\frac{n(n+1)}2(n-1)!=C\frac{(n+1)!}2$. The sum of $0$, $1$, $\dots$, $n!-1$ is $\frac{(n!-1)n!}{2}$. Since $n$ is odd, $n!|\frac{(n+1)!}2$ but since $n!-1$ is odd, $n!\not|\frac{(n!-1)n!}2$, which means our original statement is proven, as desired.

Assume that all the $S(a)$ are distinct $\pmod{n!}$. Then they must cover all residues $\pmod{n!}$. Now,consider $\mathcal{S} = \sum_{a \in [n]} S(a)$. Note that $\mathcal{S} \equiv 1 + \dots + n! \equiv \frac{(n!)(n!+1)}{2} \equiv \frac{n!}{2} \pmod{n!}.$ But we can also write $\mathcal{S}$ as $$\sum_{i = 1}^n\sum_{a \in [n]} \sum c_ia_i$$$$\equiv (n-1)!c_i \sum_{i = 1}^n i$$$$\equiv 0 \pmod{n!},$$contradiction. $\square$

Suppose to the contrary that all $n!$ values of $S(a)$ are distinct mod $n!$. Summing $S$ over all permutations, we have \[\sum_{a \colon [n] \rightarrow [n]} S(a) = \sum_{i = 1}^n ((n - 1)!(1 + 2 + \dots + n)c_i) = \frac{n + 1}{2} \cdot n! \cdot (c_1 + c_2 + \dots + c_n),\]which is divisible by $n!$ since $n$ is odd. But by assumption \[\sum_{a \colon [n] \rightarrow [n]} S(a) \equiv 0 + 1 + \dots + (n! - 1) = \frac{n!(n! - 1)}{2} \equiv \frac{n!}{2} \pmod{n!},\]a contradiction. $\square$

If this were not true, all the permutations would have distinct $S$ values mod $n!$, implying that their total sum mod $n!$ would be $\frac 12 n! (n! - 1 ) \equiv \frac 12 n! \mod n!$, but we also have $\sum S(p) = (n - 1)! n(n + 1) \frac 12 (\sum c_i) = n! (\frac{n + 1}{2}) \sum c_i \equiv 0 \mod n! $ where the sum is over all permutations $p$, which is a contradiction.

For the sake of a contradiction, suppose otherwise. Then each $S(k)$ has a different residue mod $n!,$ so the sum of all the $S_i$ is $$\frac{n!(n!-1)}{2} \pmod {n!}.$$However, we can also count this as $$(n-1)!\left( \sum^n_{i=1} c_i \right) \left( \sum^n_{i=1} a_i \right) = \frac{(n+1)!}{2} \sum^n_{i=1} c_i \pmod {n!}.$$However, note that since $(n+1)$ is even, $n!$ divides $\frac{(n+1)!}{2}.$ However, $n!-1$ is odd so $n!$ cannot possibly divide $\frac{n!(n!-1)}{2},$ a contradiction. QED

Contradiction and double counting finsihed

Assume the contrary, thus the $S(a)$ must biject to residues mod $n!$, namely $\{1, 2, \dots, n!\}$. Adding up all $S(a)$ gives $\frac{n(n+1)}{2} (\sum c_i) (n-1)!$ but this is also equal to $\frac{n!(n!+1)}{2}$ mod $n!$. The left side is clearly zero mod $n!$ but the right side is not because $n!+1$ is odd, done.

Suppose, FTSOC, there doesn't. That means all the permutations form a complete residue system $\mod{n!}$. Summing up all the permutations, we see that: $$c_1((n - 1)! + (n - 1)! \cdot 2 + ... + (n - 1)! \cdot n) + ... + c_n((n - 1)! + (n - 1)! \cdot 2 + ... + (n - 1)! \equiv \frac{n!(n! + 1)}{2} \mod{n!}$$That means $\frac{n!(n! + 1)}{2} \equiv 0\mod{n!}$, a clear contradiction as $n!$ is even, so we're done.