Let $N=100$, and let $A$ be a subset of $\lbrace 1,2,3,4, \ldots, 3N\rbrace$
containing at least $N+2$ distinct elements. We have to find $x,y\in A$ with
$N \lt y-x \lt 2N$.
Let $B=A \cap \lbrace 1,2,3,4, \ldots, N+1\rbrace$ ; denote by $t$ the
number of elements in $B$, and by $b_1 \lt b_2 \lt \ldots \lt b_t$ the elements
in $B$. Suppose first that $t\neq 0$.
If one of the $N-1$ elements $b_1+(N+1),b_1+(N+2), \ldots , b_1+(2N-1)$ is also
in $A$, taking $x=b_1$ we are done.
If one of the $t-1$ elements $b_2+(2N-1), \ldots, b_t+(2N-1)$ is also in $A$, then
we may take $y$ to be that element and we are done.
Otherwise, we have just excluded $N-1+t-1$ elements from $A$, so
$$|A\cap \lbrace N+2,N+3 \ldots,3N \rbrace| \leq
|\lbrace N+2,N+3 \ldots,3N \rbrace|-(N+t-2)=(2N-1)-(N+t-2)=N-t+1.$$
The total number of elements in $A$ is then at most
So we must have $A \cap \lbrace 1,2,3,4, \ldots, N+1\rbrace=\emptyset$, and by symmetry $A \cap \lbrace 2N-1,2N, \ldots, 3N\rbrace=\emptyset$. But then
$A \subseteq \lbrace N+2,N+3, \ldots, 2N-2\rbrace$ and $A$ has at most $N-3$ elements, contradiction.