Solution:
We first show that the assertion follows from the following lemma.

LEMMA. The polynomial $P(z)=z^{2 n}+a z^{2 n-1}+a z^{2 n-2}+\cdots+a z+1$ has at least $2 n-2$ complex zeros lying on the unit circle and different from $\pm 1$.

Since the coefficients of $P(z)$ are real numbers, its non-real zeros are complex conjugate. It follows from the lemma that there are at least $n-1$ pairs of such zeros and we denote them by $\alpha_{1}, \overline{\alpha_{1}}, \ldots, \alpha_{n-1}, \overline{\alpha_{n-1}}$.

We may assume that
$$\begin{aligned} x^{2}+b_{1} x+c_{1}= & \left(x-\alpha_{1}\right)\left(x-\overline{\alpha_{1}}\right) \\ \cdots & \cdots \\ x^{2}+b_{n-1} x+c_{n-1}= & \left(x-\alpha_{n-1}\right)\left(x-\overline{\alpha_{n-1}}\right) \end{aligned}$$
Since $|\alpha_{1}|^{2}=\cdots=|\alpha_{n-1}|^{2}=1$ we get $c_{1}=\cdots=c_{n-1}=1$ and therefore $c_{n}=1$.

Proof of the Lemma. Since the non-real zeros of $P(z)$ are complex conjugate it is enough to prove that this polynomial has at least $n-1$ zeros on the upper unit semicircle.

To do this set $z=e^{i 2 \theta}$. Then
$$\begin{aligned} \frac{z^{2 n}+1}{z^{2 n-1}+\cdots+z} & =\frac{\left(z^{2 n}+1\right)(z-1)}{z\left(z^{2 n-1}-1\right)} \\ & =\frac{\left(e^{i 2 n \theta}+e^{-i 2 n \theta}\right)\left(e^{i \theta}-e^{-i \theta}\right)}{\left(e^{i(2 n-1) \theta}-e^{-i(2 n-1) \theta}\right)} \\ & =2 \frac{\cos 2 n \theta \sin \theta}{\sin (2 n-1) \theta}=\frac{\sin (2 n+1) \theta}{\sin (2 n-1) \theta}-1 \end{aligned}$$
Therefore we have to prove that the equation
$$f(\theta)=\sin (2 n+1) \theta+(a-1) \sin (2 n-1) \theta=0$$
has at least $n-1$ roots in the interval $\left(0, \frac{\pi}{2}\right)$. This is obvious for $a=1$ because $f\left(\theta_{k}\right)=0$, where $\theta_{k}=\frac{k \pi}{2 n+1}, 1 \leq k \leq n$.

Observe now that $(k-1) \pi<(2 n-1) \theta_{k}<k \pi$ and this implies that $(-1)^{k-1} \sin (2 n-1) \theta_{k}>0$. Hence $f\left(\theta_{k}\right) f\left(\theta_{k+1}\right)<0$ for $a \neq 1$ and the Intermediate value theorem implies that the equation $f(\theta)=0$ has at least one root in each of the intervals $\left(\theta_{1}, \theta_{2}\right), \ldots,\left(\theta_{n-1}, \theta_{n}\right)$. This completes the proof of the lemma.

Remark. It follows from the above that the polynomial $P(z)$ has at most two real zeros $x_{1}$ and $x_{2}$ (possibly $x_{1}=x_{2}$ ) and $x_{1} x_{2}=1$. It can be proved that:

1) if $a>2$, then $x_{1}<-1$ and $x_{2} \in(-1,0)$;

2) if $a=2$, then $x_{1}=x_{2}=-1$;

3) if $a< -\frac{2}{2 n-1}$, then $x_{1}>1$ and $x_{2} \in(0,1)$;

4) if $a=-\frac{2}{2 n-1}$, then $x_{1}=x_{2}=1$;

5) if $a \in\left(-\frac{2}{2 n-1}, 2\right)$, then the polynomial $P(z)$ has no real zeros. In this case it has a zero $z=e^{i 2 \theta}$, where $\theta \in\left(0, \theta_{1}\right)$ for $a \in\left(-\frac{2}{2 n-1}, 1\right)$ and $\theta \in\left(\theta_{n}, \frac{\pi}{2}\right)$ for $a \in(1,2)$.