256. Let all numbers $a_{1}, a_{2}, \ldots, a_{n}$ (where $n \geqslant 2$) be positive; how many real solutions does the system of equations have:

$$
x_{1} x_{2}=a_{1}, x_{2} x_{2}=a_{2}, \ldots, x_{n-1} x_{n}=a_{n-1}, x_{n} x_{1}=a_{n} ?
$$
256. Let's consider two cases separately.

$1^{\circ} . n$ is even. By multiplying the "odd" (i.e., $1$-st, $3$-rd, ..., $(n-1)$-th) and "even" equations of our system, we get:

$$
x_{1} x_{2} x_{3} \ldots x_{n}=a_{1} a_{3} a_{5} \ldots a_{n-1} \text { and } x_{1} x_{2} x_{3} \ldots x_{n}=a_{2} a_{4} a_{6} \ldots a_{n}
$$

from which it is clear that if $a_{1} a_{3} a_{5} \ldots a_{n-1} \neq a_{2} a_{4} a_{6} \ldots a_{n}$, the system has no solutions. If, however, $a_{1} a_{3} a_{5} \ldots a_{n-1}=a_{2} a_{4} a_{6} \ldots a_{n}$, then by taking any value for $x_{1}$ (certainly different from 0), we can subsequently find from the 1-st, 2-nd, ..., $(n-1)$-th equations of the system:

$$
x_{2}=\frac{a_{1}}{x_{1}}, \quad x_{3}=\frac{a_{2}}{x_{2}}, \ldots, \quad x_{n}=\frac{a_{n-1}}{x_{n-1}}
$$

Substituting all these values into the last equation shows that it is also satisfied.

$2^{\circ} . n$ is odd. Dividing the product of the "odd" (1-st, 3-rd, ..., $n$-th) equations by the product of the "even" equations, we get:

$$
x_{1}^{2}=\frac{a_{1} a_{3} a_{5} \ldots a_{n}}{a_{2} a_{4} \ldots a_{n-1}}, \quad \text { hence } x_{1}= \pm \sqrt{\frac{a_{1} a_{3} a_{5} \ldots a_{n}}{a_{2} a_{4} a_{6} \ldots a_{n-1}}}
$$

(we recall that all $a_{i}$ are positive). Then from the 1-st, 2-nd, ..., $(n-1)$-th equations, we sequentially find:

$$
x_{2}=\frac{a_{1}}{x_{1}}, \quad x_{3}=\frac{a_{2}}{x_{2}}, \ldots, \quad x_{n}=\frac{a_{n-1}}{x_{n-1}}
$$

Verification shows that, due to (*), the last equation of the system is also satisfied.

Answer: 0, if $n$ is even and $a_{1} a_{3} \ldots a_{n-1} \neq a_{2} a_{4} \ldots a_{n}$; infinitely many, if $n$ is even and $a_{1} a_{3} \ldots a_{n-1}=a_{2} a_{4} \ldots a_{n}$; 2, if $n$ is odd.