*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 $7+8 > 12$.