$12 \cdot 26$ Find positive integers $x_{1}, x_{2}, \cdots, x_{29}$, where at least one is greater than 1988, such that $x_{1}^{2}+x_{2}^{2}+\cdots+x_{29}^{2}=29 x_{1} x_{2} \cdots x_{29}$ holds.
(29th International Mathematical Olympiad Candidate Problem, 1988)
[Solution 1] Let $\left(x_{1}, x_{2}, \cdots, x_{29}\right)$ be a set of integer solutions to the equation
$$
x_{1}^{2}+x_{2}^{2}+\cdots+x_{29}^{2}=29 x_{1} x_{2} \cdots x_{29}
$$

Let
$$
\begin{array}{c}
x_{1}^{2}+x_{2}^{2}+\cdots+x_{28}^{2}=q, \\
29 x_{1} x_{2} \cdots x_{28}=p .
\end{array}
$$

Then equation (1) can be rewritten as

i.e.,
$$
\begin{aligned}
q+x_{29}^{2} & =p x_{29}, \\
x_{29}^{2}-p x_{29}+q & =0 .
\end{aligned}
$$

Thus, by (2), $x_{29}$ and $p-x_{29}$ are the two roots of the equation
$$
t^{2}-p t+q=0
$$

This indicates that if $\left(x_{1}, x_{2}, \cdots, x_{29}\right)$ is a set of integer solutions to equation (1), then $\left(x_{1}, x_{2}, \cdots, x_{28}, 29 x_{1} x_{2} \cdots x_{28}-x_{29}\right)$ is also a set of integer solutions to equation (1).

Clearly, $(1,1, \cdots, 1)$ is a solution to (1). By the above conclusion, $(1,1, \cdots, 1, 29 \cdot 1 \cdot 1 \cdots 1-1)=(1,1, \cdots, 1,28)$ is another solution to (1).
By symmetry, $(\underbrace{1,1, \cdots, 1}_{27 \text { of }}, 28,1)$ is also a solution to (1). Therefore,
$$
\begin{aligned}
& \left(1,1, \cdots, 1,28,29 \cdot 28 \cdot 1^{27}-1\right) \\
= & (1,1, \cdots, 1,28,811)
\end{aligned}
$$

is a solution to (1).
Again by symmetry, $(1,1, \cdots, 1,811,28)$ is a solution to (1), thus
$$
\begin{aligned}
& \left(1,1, \cdots, 1,811,29 \cdot 811 \cdot 1^{27}-28\right) \\
= & (1,1, \cdots, 1,811,23491)
\end{aligned}
$$

is also a solution to (1).
Therefore, $(1,1, \cdots, 1,811,23491)$ is a set of solutions that meet the problem's requirements. Additionally, the above method can yield infinitely many sets of integer solutions that meet the problem's requirements.
[Solution 2] Let $\left(x_{1}, x_{2}, \cdots, x_{29}\right)$ be a set of integer solutions to the equation
$$
x_{1}^{2}+x_{2}^{2}+\cdots+x_{29}^{2}=29 x_{1} x_{2} \cdots x_{29}
$$

To make one of them greater than 1988, consider giving $x_{1}$ an increment $h$.
For this, set $x^{\prime}{ }_{1}=x_{1}+h, h>0$.
And assume $x^{\prime}{ }_{1}, x_{2}, \cdots, x_{29}$ still satisfy equation (1), then
$$
\left(x_{1}+h\right)^{2}+x_{2}^{2}+\cdots+x_{29}^{2}=29\left(x_{1}+h\right) x_{2} \cdots x_{29} \text {. }
$$
(2) - (1) gives
$$
2 x_{1} h+h^{2}=29 h x_{2} x_{3} \cdots x_{29} \text {. }
$$

By $h \neq 0$ we get
$$
h=29 x_{2} x_{3} \cdots x_{29}-2 x_{1} .
$$

Since $\left(x_{1}, x_{2}, \cdots, x_{29}\right)=(1,1, \cdots, 1)$ is a solution to equation (1), we have
$$
h=27, x_{1}^{\prime}=28 \text {. }
$$

Clearly, $(28,1,1, \cdots, 1)$ is a solution to equation (1).
Since (3) holds for any $x_{i}$, we also have
$$
\begin{aligned}
h & =29 x^{\prime}{ }_{1} x_{3} x_{4} \cdots x_{29}-2 x_{2} \\
& =812-2 \\
& =810 .
\end{aligned}
$$

Thus, $\quad x^{\prime}{ }_{2}=x_{2}+h=811$.
Then let
$$
\begin{aligned}
h & =29 x^{\prime}{ }_{2} x_{3} \cdots x_{29}-2 x^{\prime}{ }_{1} \\
& =29 \cdot 811-2 \cdot 28 \\
& =23463 .
\end{aligned}
$$

Then
$$
x^{\prime \prime}{ }_{1}=28+23463=23491 \text {. }
$$

Thus, $(23491,811,1,1, \cdots, 1)$ is a set of solutions to equation (1). This set of solutions meets the problem's requirements.