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, the difference of whose square roots is less than 1?
Answer: 45.

Solution. We will show that $k=45$ works. Let's divide the numbers from 1 to 2016 into 44 groups:

\[
\begin{aligned}
& (1,2,3), \quad(4,5,6,7,8), \quad(9,10, \ldots, 15), \ldots, \quad\left(k^{2}, k^{2}+1, \ldots, k^{2}+2 k\right), \ldots, \\
& (1936,1937, \ldots, 2016)
\end{aligned}
\]

Since there are 45 numbers, some two of them (let's call them $a$ and $b$) will end up in the same group. Suppose for definiteness that $a<b$. Then $k^{2} \leqslant a<b<(k+1)^{2}$ and, consequently, $\sqrt{b}-\sqrt{a}<(k+1)-k=1$.

Now let's present 44 numbers, all differences between the square roots of which are at least 1:

\[
\begin{array}{llllll}
1^{2}, & 2^{2}, & 3^{2}, & 4^{2}, & \ldots, & 44^{2} .
\end{array}
\]