Answer. $160$.

If two people are not friends, we will say that they are enemies. The total number of pairs of people in this company is $\frac{20 \cdot 19}{2} = 190$, so it is enough to prove that the minimal number of pairs of enemies is $30$.

Let us prove that there cannot be fewer than $30$ pairs of enemies. Suppose the contrary. Then there is a person $A$ who has at most two enemies (if every person has at least three enemies, the number of pairs of enemies is at least $\frac{3 \cdot 20}{2} = 30$). Suppose $A$ is seated at some table $T$ in a successful arrangement. By the condition, there are exactly $5$ people at table $T$. Since $A$ has at most two enemies, at one of the remaining three tables, all the people are friends of $A$. Therefore, if we move $A$ to that table (and do not change the rest of the arrangement), the new arrangement will also be successful, but table $T$ will have only $4$ people, which contradicts the condition.

Let us give an example satisfying the condition, in which there are exactly $30$ pairs of enemies. Divide all people into $5$ groups of $4$, and let any pair of people from the same group be enemies, and any pair from different groups be friends. In this case, each person has exactly $3$ enemies, and there are $\frac{3 \cdot 20}{2} = 30$ pairs of enemies in total. In the described situation, an arrangement is successful if and only if people from the same group are seated at different tables. Thus, successful arrangements exist, and in any successful arrangement, each table will have exactly one person from each group, that is, exactly $5$ people.