Croatian Mathematical Olympiad

Let $k = 3 \cdot 4^l$ for an arbitrary positive integer $l$. Then we have
$$a_k = \left\lfloor \frac{2^k}{k} \right\rfloor = \left\lfloor \frac{2^{3 \cdot 4^l}}{3 \cdot 4^l} \right\rfloor = \left\lfloor \frac{2^{3 \cdot 4^l - 2l}}{3} \right\rfloor = \left\lfloor \frac{2^{3 \cdot 4^l - 2l} - 1}{3} + \frac{1}{3} \right\rfloor.$$
Since the positive integer $3 \cdot 4^l - 2l$ is even, we have $2^{3 \cdot 4^l - 2l} \equiv 4 \equiv 1 \pmod 3$, so the odd positive integer $2^{3 \cdot 4^l - 2l} - 1$ is divisible by $3$.
Therefore,
$$a_k = \left\lfloor \frac{2^k}{k} \right\rfloor = \left\lfloor \frac{2^{3 \cdot 4^l - 2l} - 1}{3} + \frac{1}{3} \right\rfloor = \frac{2^{3 \cdot 4^l - 2l} - 1}{3}$$
is odd, which finishes the proof.