2. On 2016 cards, numbers from 1 to 2016 were written (each one once). Then $k$ cards were taken. What is the smallest $k$ such that among them there will be two cards with numbers $a$ and $b$ such that $|\sqrt[3]{a}-\sqrt[3]{b}|<1$?
Answer: 13.

Solution. We will show that $k=13$ works. Divide the numbers from 1 to 2016 into 12 groups:

\[
\begin{aligned}
& (1,2,3,4,5,6,7), \quad(8,9, \ldots, 26), \quad(27,28, \ldots, 63), \ldots \\
& \left(k^{3}, k^{3}+1, \ldots,(k+1)^{3}-1\right), \ldots,(1728,1729, \ldots, 2016)
\end{aligned}
\]

Since there are 13 numbers, some two of them (let's call them $a$ and $b$) will end up in the same group. Let's assume for definiteness that $a<b$. Then $k^{3} \leqslant a<b<(k+1)^{3}$ and, consequently, $0<\sqrt[3]{b}-\sqrt[3]{a}<(k+1)-k=1$.

Now let's present 12 numbers, all differences between the cube roots of which are at least 1:

\[
1^{3}, \quad 2^{3}, \quad 3^{3}, \quad 4^{3}, \quad \ldots, \quad 12^{3} .
\]