Determine the value of the natural number $a$, knowing that $4 a^{2}$ and $\frac{4}{3} \times a^{3}$ are four-digit integers.
We have $1000 \leq 4 a^{2}<10000$, which implies $250 \leq a^{2}<2500$. But, $15^{2}=225,16^{2}=256$ and $50^{2}=2500$, therefore, since $a$ is a natural number, we get $15<a<50$. We also have $1000 \leq \frac{4}{3} \times a^{3}<10000$, which implies $750 \leq a^{3}<7500$. But, $9^{3}=729,10^{3}=1000,19^{3}=6859$ and $20^{3}=8000$, therefore, since $a$ is a natural number, we also have $9<a<20$. Thus, the only options are $a=16,17,18$ or 19.

On the other hand, since $\frac{4}{3} \times a^{3}$ is an integer, we conclude that $a=18$. Indeed, this can be obtained by substituting the four possible values of $a$ or, alternatively, by observing that $a^{3}$ must be a multiple of $3$ and, consequently, that $a$ is a multiple of 3.