Estonian Math Competitions

Answer: $\{5,20\},\{5,6,14\},\{5,7,13\},\{5,8,12\},\{5,9,11\},\{6,12\}$.

Clearly there must be at least 2 numbers. Let $n$ be the smallest number, $s$ the sum and $k$ the product of the remaining numbers. We get the equation $4\cdot \frac{n+s}{nk}=\frac{s}{k}$. Multiplying by $nk$, we get $4(n+s)=ns$, which rearranges to $(n-4)(s-4)=16$. We know that $16=1\cdot 16=2\cdot 8=4\cdot 4=8\cdot 2=16\cdot 1$ and $n<s$, as $n$ was the smallest number on the blackboard. This leaves the options $n-4=1,s-4=16$ and $n-4=2,s-4=8$ (note that negative factors would also yield negative $n$ and/or $s$). So $n=5$ and $s=20$ or $n=6$ and $s=12$.

Let $n=5$ and $s=20$. If there are initially 2 numbers, they must be $5$ and $20$. If there are 3 numbers, they could be $5,6$ and $14$ or $5,7$ and $13$ or $5,8$ and $12$ or $5,9$ and $11$. There can't be 4 or more numbers, because $6+7+8>20$.

Let $n=6$ and $s=12$. If there are initially 2 numbers, they must be $6$ and $12$. There can't be 3 or more numbers, because $7+8>12$.