, 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.