Solution:

Answer: $20$

To minimize the sum, we want each player to say an estimate as small as possible—i.e., an estimate as close to $80\%$ of his actual number of cards as possible. We claim that the minimum possible sum is $20$.

First, this is achievable when $R2$ has $10$ cards and estimates $8$, and when $R3$ has $14$ cards and estimates $12$.

Then, suppose that $R2$ has $x$ cards and $R3$ has $24-x$. Then, the sum of their estimates is
$$\left\lceil \frac{4}{5}x \right\rceil + \left\lceil \frac{4}{5}(24-x) \right\rceil \geq \left\lceil \frac{4}{5}x + \frac{4}{5}(24-x) \right\rceil = \left\lceil \frac{4}{5} \times 24 \right\rceil = \left\lceil 19.2 \right\rceil = 20$$
Note: We use the fact that for all real numbers $a, b$, $\lceil a \rceil + \lceil b \rceil \geq \lceil a + b \rceil$.