Let $S$ be the set of all real numbers that can be expressed as $\pm t^n \pm t^{t-1} \pm \cdots \pm 1$. We will prove that for all $d\geq 1, S\cap [d,d+2] \neq \varnothing$. Indeed, assume the statement is true for $d$. We will prove it for $td+1$. Indeed, take any $x\in S\cap [d,d+2]$. If $x\in [d,d+2/t]$ then $tx + 1\in [td + 1, td + 3]$. Otherwise, $tx - 1 \in [td + 1, td + 2t - 1]\subset [td+1,td+3]$. Notice that $tx -1$ and $tx+1$ also belong to $S$. Thus, if the statement is true for $d$ then it is true for $kd+1$. Since $\{1,t+1\}\subset S$, it implies that $S\cap [d,d+2] \neq \varnothing,\forall 1\leq d\leq t+1\Rightarrow S\cap [d,d+2],\forall d\geq 1$. Hence, $S\cap [2018, 2020]$. (q.e.d)