Example 6 Find all positive integers $n \geqslant 2$, such that the system of equations
$$\left\{\begin{array}{c}
x_{1}^{2}+x_{2}^{2}+50=16 x_{1}+12 x_{2} \\
x_{2}^{2}+x_{3}^{2}+50=16 x_{2}+12 x_{3} \\
\cdots \\
x_{n-1}^{2}+x_{n}^{2}+50=16 x_{n-1}+12 x_{n} \\
x_{n}^{2}+x_{1}^{2}+50=16 x_{n}+12 x_{1}
\end{array}\right.$$

has integer solutions.
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.