Iranian Mathematical Olympiad

At first we shall prove following lemma:
Lemma. Let $b_n$ be a sequence of positive real numbers satisfying
$$b_n \leq 2b_0 + b_1 + \cdots + b_{n-1},$$
then for each real number $z$ where
$$|z| \leq 2b_0 + b_1 + \cdots + b_n,$$
there are $a_0, a_1, \dots, a_n \in \{1, -1\}$ such that
$$\left|z - \sum_{i=0}^{n} a_i b_i\right| < b_0.$$
Proof. Write the inequality $|z| \leq 2b_0 + b_1 + \cdots + b_n$ in the form
$$|z - \operatorname{sgn}(z)b_n| \leq 2b_0 + b_1 + \cdots + b_{n-1},$$
then, proceed the proof by induction on $n$. For sake of convenience, we also define $\operatorname{sgn}(0) = 1$. This completes our proof. $\square$

Back to the problem, define $b_i = t^i$, then it is easy to deduce that
$$b_n - b_0 = t^n - 1 \le 1 + t + \cdots + t^{n-1} = \frac{t^n - 1}{t-1} = b_0 + b_1 + \cdots + b_{n-1}.$$
Moreover, choose $d$ such that $t^d \ge 2019$, then
$$2019 \le t^d \le 2 + t + \cdots + t^d.$$
Hence, by the lemma, there are $a_0, a_1, \dots, a_d \in \{1, -1\}$ such that
$$\left| \sum_{i=0}^{d} a_i t^i - 2019 \right| < 1.$$
We are done.