Let $M$ be the set of the numbers $n$ that satisfy the conditions above. We shall show that $M = \mathbb{N} \setminus \{0, 1, 3\}$.

We can observe that $2p \in M$, for any $p \ge 1$, the relations being satisfied e.g. for $x_1 = x_2 = \dots = x_p = 1$ and $x_{p+1} = x_{p+2} = \dots = x_{2p} = -1$.

Moreover, if $n \in M$ and $x_1 + x_2 + \dots + x_n = \frac{1}{x_1} + \frac{1}{x_2} + \dots + \frac{1}{x_n} = 0$, taking $x_{n+1} = 1$ and $x_{n+2} = -1$, we conclude that $n+2 \in M$, because
$$x_1 + x_2 + \dots + x_n + x_{n+1} + x_{n+2} = \frac{1}{x_1} + \frac{1}{x_2} + \dots + \frac{1}{x_n} + \frac{1}{x_{n+1}} + \frac{1}{x_{n+2}} = 0.$$

Therefore, all that is left to be done is to find the smallest odd element of $M$.
Suppose that $3 \in M$; then there exist $x_1, x_2, x_3 \in \mathbb{R}^*$ so that
$$x_1 + x_2 + x_3 = \frac{1}{x_1} + \frac{1}{x_2} + \frac{1}{x_3} = 0.$$

This leads to $x_1^2 + x_1 x_2 + x_2^2 = 0$, which is possible if and only if $x_1 = x_2 = 0$, contradiction. It follows that $3 \notin M$.

Let us search for an example to prove that $5 \in M$. Take $x_1 = x_2 = x_3 = -1$; then $x_4 + x_5 = 3$ and $x_4 x_5 = 1$, which leads to $x_{4,5} = \frac{3 \pm \sqrt{5}}{2}$.