Solving by rearranging and completing the square, the system of equations transforms to
$$\left\{\begin{array}{c} \left(x_{1}-8\right)^{2}+\left(x_{2}-6\right)^{2}=50 \\ \left(x_{2}-8\right)^{2}+\left(x_{3}-6\right)^{2}=50 \\ \cdots \\ \left(x_{n-1}-8\right)^{2}+\left(x_{n}-6\right)^{2}=50 \\ \left(x_{n}-8\right)^{2}+\left(x_{1}-6\right)^{2}=50 \end{array}\right.$$

Since 50 can be expressed as the sum of squares of two positive integers in only two ways: $50=1^{2}+7^{2}=5^{2}+5^{2}$, it follows from (1) that $\left|x_{2}-6\right|=1, 5$ or 7, and from (2) that $\left|x_{2}-8\right|=1, 5$ or 7, thus $x_{2}=1, 7$ or 13.

Furthermore, for each $1 \leqslant i \leqslant n$, we have $x_{i}=1, 7$ or 13. Depending on $x_{1}=1, 7, 13$, we consider three cases.

If $x_{1}=1$, then from (1) we know $x_{2}=7$, and from (2) we know $x_{3}=13$, and so on. Thus, when $k \equiv 1(\bmod 3)$, $x_{k}=1$; when $k \equiv 2(\bmod 3)$, $x_{k}=7$; and when $k \equiv 0(\bmod 3)$, $x_{k}=13$. Therefore, by the second equation, we know that the original system of equations has integer solutions if and only if $n+1 \equiv 1(\bmod 3)$, i.e., if and only if $3 \mid n$, $n$ meets the requirement.

For the other two cases $x_{1}=7$ and $x_{1}=13$, similar discussions yield the same condition.
In summary, the $n$ that satisfies the condition are all multiples of 3.