National Math Olympiad in Slovenia

Let $a$ and $b$ be the lengths of the sides of a rectangle that satisfies the conditions. Then $2(a+b) = n = ab$. This equality can be rewritten as $a(b-2) = 2b$. Now we can express $a$ in terms of $b$:
$$a = \frac{2b}{b-2} = \frac{2(b-2)}{b-2} + \frac{4}{b-2} = 2 + \frac{4}{b-2}.$$

Since $a$ and $b$ are positive integers, the number $b-2$ must divide $4$. On the other hand $b-2$ is greater than $-2$, so we conclude that it can only be equal to $-1, 1, 2$ or $4$. Thus, $b$ equals $1, 3, 4$ or $6$ and $a$, respectively, equals $-2, 6, 4$ or $3$. The first case is not a valid solution since $-2$ is not a positive integer. In the second and fourth case we get $n=18$, in the third case $n=16$.