An odd coincidence?

Let us denote the odd integer $n$ as $O$ . $P$ and $S$ as product and sum of $n$ respectively and we have $\begin{aligned} & (P/S)_{n\in\mathbb O}=\dfrac{1\times 2\times3 \times \cdots n}{1+2+ 3+\cdots+ n } = \dfrac{n!}{\frac{n(n+1)}{2}} \\& (P/S)_{n\in\mathbb O}=\dfrac{2n!}{n(n+1)} = \frac{2(n-1)!}{n+1} \end{aligned}$ We can cleary note that $(n-1)$ and $(n+1)$ are even integers since $n$ is an odd integer that implies that $(n-1)! = (2m_1)!$ and $n+1 = 2m_2$ where $(m_1,m_2)\in\mathbb N$ also $m_2-m_1 =2$ $\begin{aligned}& (P/S)_{n\in\mathbb O} = \dfrac{2(n-1)!}{n+1} = \frac{2(2m_1)!}{2m_2}\\& (P/S)_{n\in\mathbb O} =\dfrac{(2m_1)!}{m_2}= \dfrac{(2(m_2-2))!}{m_2}\end{aligned}$ which is always divisible. Hence the answer is Yes .

We know that $S(n) = \displaystyle\sum^{n}_{k = 1}k = \dfrac{n(n+1)}{2}$ .

If we now divide $n!$ by $S(n)$ as such, $\dfrac{n!}{S(n)} = \dfrac{2n!}{n(n+1)} = \dfrac{2(n - 1)!}{n+1}$ . Note that $n$ is an odd integer and therefore $n + 1$ is even. Because of this, we can express $n + 1$ as $2x$ where $x \in \mathbb{Z}^+$ and therefore we can simplify our fraction to $\dfrac{(n - 1)!}{x}$ and since $x < (n - 1)$ it is contained within $(n - 1)!$ and thus will divide into it. Therefore the answer is $\boxed{\text{Yes}}$ .

Much easier solution that others.

The product will be equal to n! , then sum 1+2+..+n = n(n+1)/2. As n is odd (n+1) will be even, and when divided by 2 it, number comes less than 'n' and let it be 'x'.

Now n!/(n \times x), it will be always divisible.