Brazilian Math Olympiad

We will choose two large positive integers $r$, $s$, and take $m = r + s$, $a_j = 1$ for $1 \le j \le r$ and $a_j = 2$ for $r + 1 \le j \le r + s = m$.
We have $q_{k+1} = q_k + q_{k-1}$, for $1 \le k \le r-1$, and so $q_j = F_{j+1}$, for $0 \le j \le r$, where
$$F_j = \frac{1}{\sqrt{5}} \left( \left( \frac{1+\sqrt{5}}{2} \right)^j - \left( \frac{1-\sqrt{5}}{2} \right)^j \right) = \frac{1+o(1)}{\sqrt{5}} \left( \frac{1+\sqrt{5}}{2} \right)^j$$
is the $j$-th term of Fibonacci's sequence, for $j \ge 1$. So $q_j = \frac{1+o(1)}{\sqrt{5}} (\frac{1+\sqrt{5}}{2})^{j+1}$ for large $j$.

On the other hand, we have $q_{k+1} = 2q_k + q_{k-1}$ for $r \le k \le m-1$, and so $q_{r+j} = u_{j+1}q_r + u_j q_{r-1}$, where $(u_j)_{j \ge 0}$ is the sequence given by $u_0 = 0$, $u_1 = 1$ and $u_{k+2} = 2u_{k+1} + u_k$, for $k \ge 0$. Since
$$u_k = \frac{1}{2\sqrt{2}}((1+\sqrt{2})^k - (1-\sqrt{2})^k) = \frac{1+o(1)}{2\sqrt{2}}(1+\sqrt{2})^k, \quad \text{for } k \ge 0,$$
we get
$$\begin{align*} q_{r+j} &= \frac{1+o(1)}{2\sqrt{2}}((1+\sqrt{2})q_r + q_{r-1})(1+\sqrt{2})^j \\ &= \frac{1+o(1)}{2\sqrt{10}}\left(1+\sqrt{2}+\frac{\sqrt{5}-1}{2}\right)(1+\sqrt{2})^j\left(\frac{1+\sqrt{5}}{2}\right)^{r+1} \\ &= (1+o(1))\frac{4+\sqrt{10}+\sqrt{2}}{8\sqrt{5}}(1+\sqrt{2})^j\left(\frac{1+\sqrt{5}}{2}\right)^{r+1}, \end{align*}$$
provided that $j$ and $r$ are large.

Since $\log(1 + \sqrt{2}) / \log(\frac{1+\sqrt{5}}{2})$ is irrational, the result follows (by taking logarithms) from the elementary fact below:
Given $\alpha$, $\beta > 0$ such that $\alpha/\beta$ is irrational, $\epsilon > 0$ and $r > 0$, there is $x_0 > 0$ such that, for every $x \in \mathbb{R}$, $x \ge x_0$, there are positive integers $m$, $n \ge r$ such that $|m\alpha + n\beta - x| < \epsilon$.