What is the probability that some people are chosen in different groups? Let's say that I have $n$ people and $m \le n$ of this people are malicious. Then I run a process that chooses 6 people randomly and assigns each of them to one of two different groups $A$ and $B$ such that each group in the end has 3 people. What is the probability that there will be at least a malicious person in every group?

Not sure if this is the correct answer, but maybe my reasoning can help you nonetheless.
The total number of arrangements is: ${n \choose 6} \cdot { 6 \choose 3}$.
${n \choose 6}$ represents the total ways of choosing 6 people out of the n people
${ 6 \choose 3}$ represents the number of ways of splitting the people into two groups
Now I will (attempt) to find the number of event in which one of the groups has no malicious persons...which is the complement of what you require.
This requires 4 separate cases:
Case 1: None of the chosen 6 people are malicious
${n-m \choose 6}\cdot {6 \choose 3}$
Case 2: one person is malicious
$2\cdot{m \choose 1} \cdot {n-(m-1) \choose 5} \cdot {5 \choose 3}$
Case 3: Two persons are malicious and placed in the same group
$2\cdot{m \choose 2} \cdot {n-(m-2) \choose 4} \cdot {4 \choose 3}$
Case 4: Three persons are malicious and placed in the same group
$2\cdot{m \choose 3} \cdot {n-(m-3) \choose 3} \cdot {3 \choose 3}$
So the final answer is: $$1-\dfrac{{n-m \choose 6}\cdot {6 \choose 3}}{{n-m \choose 6}\cdot {6 \choose 3} + 2\cdot{m \choose 1} \cdot {n-(m-1) \choose 5} \cdot {5 \choose 3} + 2\cdot{m \choose 2} \cdot {n-(m-2) \choose 4} \cdot {4 \choose 3} + 2\cdot{m \choose 3} \cdot {n-(m-3) \choose 3} \cdot {3 \choose 3}}$$
