Perhaps you mean that $ \{1,2, \ldots, n\}$ can be partitioned into two such non-empty subsets $ P_n$ and $ S_n$? For example, $ 1 \times 2 \times 4 = 3 + 5$ is a partitioning of $ \{1,2,3,4,5\}$ $ 1 \times 2 \times 6 = 3 + 4 + 5$ is a partitioning of $ \{1,2,3,4,5,6\}$ $ 1 \times 3 \times 6 = 2 + 4 + 5 + 7$ is a partitioning of $ \{1,2,3,4,5,6,7\}$ I can check if this holds for all $ n \geq 5$ but I'd rather first make sure this is actually the problem. EDIT: Yes, this works for all $ n \geq 5$. For $ n=2k$ take $ P_{2k} = \{1, k-1, 2k\}$ and for $ n=2k+1$ take $ P_{2k+1} = \{1, k, 2k\}$.

http://www.mathlinks.ro/viewtopic.php?p=131398 I just wrote it here. I don't know are there mistakes in the problem.

Well, so now you do know (and that link doesn't work, but no matter).

Sorry for the wrong link: http://www.mathlinks.ro/viewtopic.php?t=212129

When $n$ is odd we have: \begin{align*} &2+3+\dots+[ \frac{n-1}{2}-1]+[\frac{n-1}{2}+1]+...+(n-2)+n\\ &=\frac{n(n+1)}{2}-\frac{(n-1)}{2}-(n-1)-1\\ &=\frac{n^2+n-n+1-2n+2-2}{2}\\ &=1.(n-1).\frac{n-1}{2}\\ \end{align*}When $n$ is even we have, \begin{align*} &2+3+4+\dots +\dots+[\frac{n}{2}-2]+[\frac{n}{2}]+\dots+(n-1)\\ &=\frac{n(n+1)}{2}-1-[\frac{n}{2}-1]-n\\ &=\frac{n^2+n-2-n+2-2n}{2}\\ &=\frac{n(n-2)}{2}\\ &=1.\frac{n-2}{2}.n\\ &=1.[\frac{n}{2}-1].n\\ \end{align*}