50th Mathematical Olympiad in Ukraine, Third Round (January 23, 2010)

Answer: $n\ge 3$.

If $6n\le n^2$ or equivalently $n\ge 6$, then after putting $a=n$, $c=2n$, $d=3n$, $b=6n$ we will have $ab=cd$. So every $n\ge 6$ satisfies the statement.

In the case $n=5$ we have numbers $5,6,7,\dots ,25$. Considering $a=6$, $b=20$, $c=8$, $d=15$ we have the desired result.

In the case $n=4$ we have numbers $4,5,6,\dots ,16$. Considering $a=4$, $b=15$, $c=6$, $d=10$ we have the desired result.

In the case $n=3$ we can put $a=3$, $b=8$, $c=4$, $d=6$.

In the case $n=1,2$ in the set $n,n+1,n+2,\dots ,n^2$ at most three numbers, so we can't find four pairwise distinct numbers.