. Four (50 points) Find all positive integers $a$, such that there exist integer-coefficient polynomials $P(x)$ and
$$
Q(x)=a_{1} x+a_{2} x^{2}+\cdots+a_{2016} x^{2016},
$$

where, $a_{i} \in\{-1,1\}, i=1,2, \cdots, 2016$, satisfying
$$
\left(x^{2}+a x+1\right) P(x)=Q(x) \text {. }
$$
First, prove that $Q(x)=0$ has no complex roots with modulus greater than or equal to 2.

Assume $Q(x)=0$ has a complex root with modulus greater than or equal to 2, denoted as $z(|z| \geqslant 2)$.
$$
\begin{array}{l}
\text { Then } a_{1} z+a_{2} z^{2}+\cdots+a_{2016} z^{2016}=0 \\
\Rightarrow a_{1} z+a_{2} z^{2}+\cdots+a_{2015} z^{2015}=-a_{2016} z^{2016} \\
\Rightarrow\left|a_{1} z+a_{2} z^{2}+\cdots+a_{2015} z^{2015}\right| \\
\quad=\left|z^{2016}\right|=|z|^{2016} \\
\quad \leqslant|z|+\left|z^{2}\right|+\cdots+\left|z^{2015}\right|=\frac{|z|^{2016}-|z|}{|z|-1} \\
\quad \leqslant|z|^{2016}-|z|,
\end{array}
$$

Contradiction.
Thus, the proposition holds.
Since $\left(x^{2}+a x+1\right) P(x)=Q(x)$, then $\left(x^{2}+a x+1\right) \mid Q(x)$.
Therefore, $x^{2}+a x+1=0$ also has no complex roots with modulus greater than or equal to 2.
Thus, $a$ can only be 1 or 2.
When $a=1$, we can construct
$$
\begin{array}{l}
\left(x^{2}+x+1\right)\left(x^{2014}+x^{2011}+\cdots+x\right) \\
=x^{2016}+x^{2015}+\cdots+x ;
\end{array}
$$

When $a=2$, we can construct
$$
\begin{array}{l}
\left(x^{2}+2 x+1\right)(x-1)\left(x^{2013}+x^{2009}+\cdots+x\right) \\
=x^{2016}+x^{2015}-x^{2014}-x^{2013}+\cdots+x^{4}+x^{3}-x^{2}-x .
\end{array}
$$

In conclusion, the possible values of $a$ are 1 or 2.