We try $n = 5$ and get that $1, 2, 4$ multiply to the sum of $3, 5$. So we try $1, a$ and $1, a, b$. The second idea works marvelously.

Suppose we had $1, a, b$ as the multiplying guys. Then we need $ab = \frac{n(n + 1)}{2} - a - b - 1$ or equivalently $(a + 1)(b + 1) = \frac{n(n + 1)}{2}$. Now if $n$ is odd take $a = n - 1, b = \frac{n - 1}{2}$. And if $n$ is even take $a = n, b = \frac{n}{2} - 1$. This clearly works as long as $n - 1 \not= \frac{n - 1}{2} \not= 1 \not= n - 1$ for odd $n$, so when $n > 3$ is odd. And it works as long as $n \not= \frac{n}{2} - 1 \not= 1 \not= n$ for even $n$, so when $n > 4$ is even. Thus all $n \ge 5$ work. As a side note, so do $n = 1, 3$ with $\{1\}$ as the additive set and $\{\}$ as the multiplicative set for $n = 1$, and with $\{1, 2\}$ as the additive set and $\{3\}$ as the multiplicative set for $n = 3$. The others fail. Unless groups are meant to be nonempty.

Let's consider two different cases for $n$, even and odd. For each one, we will prove that the statement is true. $\textbf{Case 1}$: $n \equiv 0 \pmod 2$ Since $n$ is even we can write it as $2k$ where $k$ is an integer and $k \geq 3$. We can express the sum numbers from 1 to 2k as: $$\frac{(2k)(2k+1)}{2} = k(2k+1) = 2k^2+k$$ Now we need to find some numbers which multiply to $2k^2 +k$ minus the sum of those numbers. We claim that these numbers are always $1$, $k-1$, and $2k$ for an even value of $n$. Now, let us put that into an equation where we are comparing the sum and the product. If the numbers are $1$, $k-1$, and $2k$ then: \begin{align*} 2k^2 + k - (1) - (k-1) - (2k) &= 1(k-1)(2k) \\ 2k^2 + k - 1 - k+1 - 2k &= (k-1)(2k) \\ 2k^2 - 2k &= 2k^2 - 2k \\ \end{align*}Since these two values are equal, we have proved that for even $n$, it is possible for the sequence to be split up into two groups so that one product is equal to the other sum. $\textbf{Case 2}$: $n \equiv 1 \pmod 2$ We can write $n$ as $2k+1$ where $k \geq 2$. To find the sum of the numbers from $1$ to $2k+1$ we can do the following: $$\frac{(2k+1)(2k+2)}{2} = (2k+1)(k+1) = 2k^2 + 3k +1$$ Once again we need to find numbers that multiply to $2k^2+3k+1$ minus the sum of those numbers. In this case, we claim that these numbers can be $1$, $k$, and $2k$. Once again let us put these algebraic terms to test if these numbers work. \begin{align*} 2k^2 + 3k + 1 - (1) - (k) - 2k &= 1(k)(2k) \\ 2k^2 + 0 + 0 &= 2k^2 \\ 2k^2 &= 2k^2 \\ \end{align*} This proves the statement for the odd case of n. Since we have proved the case of $n$ being odd and also the case of $n$ being even, we have proved it for all integers where $n \geq 5$.