Croatian Mathematical Society Competitions

Denote by $S$ the set of $k+2$ given numbers. Without loss of generality, we can assume that $S$ contains $1$. Indeed, if $1$ is not in $S$, we can subtract the smallest element of $S$ from all elements of $S$ and add $1$ to all of them, which preserves the differences between all elements of $S$.

If at least one number from $\{k+2,k+3,\dots ,2k\}$ is contained in $S$, then $1$ and that number satisfy the claim.

Now assume that none of the numbers $k+2,k+3,\dots ,2k$ are in $S$. All remaining numbers from $2$ to $3k+1$ can be divided into $k$ pairs $(2,2k+1),\dots ,(k+1,3k)$. Other than $1$, the set $S$ contains $k+1$ other numbers, so by the Dirichlet principle at least one pair consists of numbers that are both in $S$. Those two numbers satisfy the claim.