Here is an interesting problem :
Is there a sequence $(\varepsilon _n)_{n \geqslant 0} \in > \{-1,1\}^{\mathbb{N}}$ such that for every $N \geqslant 1$, the polynomial $\displaystyle \sum_{n=0}^N \varepsilon _n X^n$ have $N$ real roots.
The result is true where one search for a real sequence $(\varepsilon _n)_{n \geqslant 0}$. In my opinion such a sequence with values $\pm 1$ should not exist, but I am not able to prove it.
Let $$f(x) = \sum_{n=0}^N \varepsilon_nx^n$$ be a polynomial whose roots $r_1, \dots , r_N$ are all real. WLOG we can suppose $\varepsilon_N = 1$ (i.e. $f$ is monic), and that $N \ge 2$. I will show that $N \le 3$. This will be enough to conclude that your sequence cannot exist.
First, by Vieta's formulas we have $$0 < \sum_j r_j^2 = (-\varepsilon_{N-1})^2 - 2\varepsilon_{N-2} = 1- 2\varepsilon_{N-2}$$ hence necessarily $\varepsilon_{N-2} = -1$ and $\sum_j r_j^2 = (-\varepsilon_{N-1})^2 = 1-2(-1)=3$.
The second step is to notice that $$\prod_j r_j^2 = \varepsilon_0^2 = 1$$ Applying the AM-GM inequality to the positive real numbers $r_1^2, \dots , r_N^2$ you get $$\frac{3}{N}= \frac{1}{N} \sum_j r_j^2 \ge \sqrt[N]{\prod_j r_j^2} = 1$$ so that $N \le 3$.