jmc

If $x^n+x^{n-1}+\cdots +x+1=0,$ then $$(x-1)(x^n+x^{n-1}+\cdots +x+1)=0,$$which expands as $x^{n+1}-1=0.$ Then $x^{n+1}=1.$ The only possible real roots of this equation are $x=1$ and $x=-1.$

Note that $x=1$ cannot be a real root of $$x^n+x^{n-1}+\cdots +x+1=0,$$but $x=-1$ is a root whenever $n$ is odd. Therefore, the maximum number of real roots is $\boxed{1}.$