Berkeley Math Circle

Solution:
The answer is odd $n$ only. For such odd $n$, one can take $X = -1$ as a real root.

Suppose $n$ is even. We claim $P$ has no real roots. Indeed, note first $P(1) = n + 1$, so $1$ is not a real root. But for any $x \neq 1$, we have
$$P(x) = x^{n} + \cdots + 1 = \frac{x^{n+1} - 1}{x - 1} \neq 0$$
since $n + 1$ is odd implies $x^{n+1} \neq 1$.