Find all complex-coefficient polynomials $P(x)$ such that for any three integers $a, b, c$ whose sum is not zero, the following holds:
$$
\frac{P(a)+P(b)+P(c)}{a+b+c}
$$

is an integer.
Analysis and Solution The condition given in the problem, "for any three integers $a, b, c$ whose sum is not 0, the algebraic expression (1) is an integer," is extremely useful and should be fully utilized.
First, take integers $a \neq 0, b=c=0$, we get
$$
a \mid(P(a)+2 P(0)) \text {. }
$$

Next, take integers $a=b=c \neq 0$, we get
$a \mid P(a)$.
From equations (2) and (3), we have
$a \mid 2 P(0)$
for any non-zero integer $a$. Therefore,
$$
P(0)=0 \text {. }
$$

Take integers $a, b, c$ whose sum is not 0, then the sum of $a, b+c, 0$ is also not 0. Therefore (using equation (4),
$$
\begin{array}{l}
{[a+(b+c)] \mid(P(a)+P(b+c)) .} \\
\text { Also, }(a+b+c) \mid(P(a)+P(b)+P(c)),
\end{array}
$$

From equations (5) and (6), we get
$$
(a+b+c) \mid(P(b+c)-P(b)-P(c)) \text {. }
$$

For fixed $b, c$, $a$ can be chosen arbitrarily (as long as it is not equal to $-(b+c)$), thus, equation (7) implies that for any integers $b, c$, we have
$$
P(b+c)=P(b)+P(c) .
$$

Taking $a=b=c=1$, we get $P(1)$ is an integer.
Let $P(1)=k$. Then for positive integer $n$ we have
$$
\begin{array}{l}
P(n)=P(1)+P(n-1) \\
=2 P(1)+P(n-2) \\
=\cdots=n P(1)=n k .
\end{array}
$$

By the polynomial identity theorem, we get
$$
P(x)=k x(k \in \mathbf{Z}) .
$$