If it was possible, product of the $n$ roots would $\in(0,1)$, impossible since $Q(0)\in\mathbb Z$ and $Q(x)$ monic

Trivial when $n=1$ (choose for example $Q(x)=x$) If $n\ge 2$, consider $Q(x)=x^n-3\prod_{k=0}^{n-2}(2(n-1)x-2k)+(-1)^n$ $Q(x)$ is indeed a monic polynomial $\in\mathbb Z[x]$ with degree $n$ Let $R(x)=\prod_{k=0}^{n-2}(2(n-1)x-2k)$ polynomial $\in\mathbb Z[x]$ with degree $n-1$ so that $Q(x)=x^n-3R(x)+(-1)^n$ $R(0)=0$ and so $Q(0)=(-1)^n$ Let $i\in\{1,2,...,n-1\}$ and $x_i=\frac{2i-1}{2n-2}\in(0,1)$ $R(x_i)=\prod_{k=0}^{n-2}(2i-1-2k)$ is a nonzero integer with sign of $(-1)^{n-1-i}$ So $-3R(x_i)$ has sign of $(-1)^{n-i}$ and absolute value $\ge 3$ And since $x_i^n+(-1)^n\in((-1)^n,(-1)^n+1)\subset(-1,2)$, $Q(x_i)$ is nonzero and has sign of $(-1)^{n-i}$ So we got : $Q(0)$ nonzero with sign of $(-1)^n$ $Q(x_1)$ nonzero with sign of $(-1)^{n-1}=-(-1)^n$ and so at least one real root in $(0,x_1)$ $Q(x_2)$ nonzero with sign of $(-1)^{n-2}=(-1)^n$ and so at least one real root in $(x_1,x_2)$ ... For even $i$, $Q(x_i)$ nonzero with sign of $(-1)^{n-i}=(-1)^n$ For odd $i$, $Q(x_i)$ nonzero with sign of $(-1)^{n-i}=-(-1)^n$ ... $Q(x_{n-1})$ nonzero with sign of $-1$ And so : At least one real root in $(0,x_1)$ At least one real root in each $(x_i,x_{i+1})$ for $i\in\{1,2,...,n-2\}$ And so at least $n-1$ real roots in $(0,1)$ Q.E.D.