2. From the numbers 1, 2, 3, ..., 2016, $k$ numbers are chosen. What is the smallest $k$ such that among the chosen numbers, there will definitely be two numbers whose difference is greater than 672 and less than 1344?
Answer: 674.

Solution. Let $n=672$. Then $2 n=1344$ and $3 n=2016$. Suppose it is possible to choose $674=n+2$ numbers such that no required pair of numbers can be found among them. Let $m$ be the smallest of the chosen numbers. Then the numbers $m+n+1, m+n+2$, $\ldots, m+2 n-1$ are not chosen. Remove them and the number $m$ from the set $\{1, \ldots, 3 n\}$, and denote the remaining set by $E$. Consider the pairs of numbers

\[
\begin{aligned}
& (1, n+2), \quad(2, n+3), \quad \ldots(m-1, m+n) \\
& (m+1, m+2 n), \quad(m+2, m+2 n+1), \quad \ldots, \quad(n+1,3 n) .
\end{aligned}
\]

There are $(m-1)+(n-m+1)=n$ such pairs. Note that the union of the left and right parts of these pairs gives the set $E$. Then any chosen number coincides with the left or right part of one of the pairs. By assumption, there are exactly $n+1$ such numbers, so there will be two of them, say $a$ and $b$, belonging to the same pair. Then their difference is $n+1$ or $2 n-1$. Thus, $a$ and $b$ satisfy the condition of the problem, which is impossible.

If the numbers $1,2,3, \ldots, 673$ are chosen, then it will not be possible to find the required two numbers.