EDIT: Err whoops Let me just say that perhaps the poster of this false solution is say my pet cat. Thanks for understanding, adamz

$n=6$ clearly works, take the ring $a,ab,b,\frac1a,\frac1{ab},\frac1b$ for reals $a,b$ such that the 6 values are distinct (infinitely many pairs exist, notably any two natural numbers greater than 1). A specific issue in your solution is that the smallest number is clearly not necessarily negative, as $ab>c$ is not necessarily true for $1>a,b>c$.

Also, since I might as well post a complete solution, it is easy to verify that no number under 6 works, by taking equations such as $b\cdot\frac1b=\frac1a$, giving $a=1$, giving $b=ab$, contradiction. (This is for $n=5$, there exist contradictions in all). Similarly, we can contradict $n=7,8$, and for $n\ge9$, we have a repeat of either the $\frac1a$ term or the $\frac1b$ term, completing the proof. Thus, only $n=6$ works.

If we have a number $x_i$ in the circumference of the circle we will call the number directly next to it in the clockwise direction $x_{i+1}$ and the number directly next to it in the counterclockwise direction $x_{i-1}$. Now we can write the following equations: $$x_i=x_{i-1}x_{i+1}$$$$x_{i-1}=x_{i-2}x_i$$$$x_{i+1}=x_ix_{i+2}$$Substituting we obtain: $$x_{i-2}x_{i+1}=1$$$$x_{i-1}x_{i+2}=1$$This implies that for any $x_k$ in the circumference $$x_kx_{k+3}=x_kx_{k-3}=1$$Notice that $x_{k-3}=x_{k+3}$ which would break the conditions set by the problem unless $n=6$ because going 3 units in the clockwise direction would take us to the same place as going 3 units in the counterclockwise direction. Therefore the only possible value for $n$ is 6. When $n=6$ we have the solution $$x_1,x_1x_2,x_2,\frac{1}{x_1},\frac{1}{x_1x_2},\frac{1}{x_2}$$where $x_1$ and $x_2$ represent 2 distinct real numbers. Thus $\framebox{n=6}$ is the only solution.