This is the same as the minimal number of edges such that in any graph exists a $ K_{h+1} $ . From Turan we get that the minimal number of pairs of opponents is $\binom{h}{2}p^2+1$ .

Does anyone have a solution for this?

Does anyone have a more detailed solution for this?

Cezar wrote: This is the same as the minimal number of edges such that in any graph exists a $ K_{h+1} $ . From Turan we get that the minimal number of pairs of opponents is $\binom{h}{2}p^2+1$ . I don't think this is right; if there were $h+1$ people, all enemies of each other, then clearly there would be no successful partition, so the upper bound on the answer is $\binom{h+1}{2}$.

Could you explain how did you come out with that please? I'm seriously struggling to solve this problem

The answer is $(h-1)\cdot \min (p,\frac{h}{2})+1$.

I think the answer is $\boxed{\min \left ((h-1)p + 1, \binom{h+1}{2}\right )}.$ We can show that $\binom{h+1}{2}$ is always an upper bound by taking a $K_{h+1}$ on any $h+1$ of the vertices. We can also see that $(h-1)p+1$ is always an upper bound by simply letting one of the members of parliament hate $(h-1)p + 1$ others. Let's now show that whenever there are less than $\min \left ((h-1)p + 1, \binom{h+1}{2}\right )$ pairs of opponents among $hp$ members of parliament, they can be split into $h$ houses of $p$ members so that no enmity relations exist within any house. We will prove this by induction on $h.$ When $h \le 2$ this is trivial, so suppose it holds for $h = t \ge 2.$ When $h = t+1$, we will use some terminology. For a partition of the members of parliament into $h$ houses of $p$ people, define its badness to be the number of pairs of enemies who are in the same house. By removing edges, we can assume WLOG that there is some partition with badness exactly $1.$ Suppose that $S_1 \cup S_2 \cup \cdots \cup S_h$ is a partition of the $hp$ members of parliament so that $|S_i| = p$ for all $1 \le i \le h$, and the only pair of enemies in the same house is $(a, b)$ for $a, b \in S_1.$ Call a set of $p$ members of parliament tranquil if no two of them are enemies. Define the value of a tranquil set to be the sum of the degrees of the members within it. Let $m$ be the greatest value of any tranquil set. Note that because there are less than $(h-1)p + 1$ enmity relations, there must exist some person $p \notin S_1$ so that $p$ is not an enemy of any of the members in $S_1$ (else, there'd be at least $(h-1)p + 1$ enmity relations, where the $+1$ comes from $a$ and $b$). Consider the $h+1$ sets of $p$ people $S_2, S_3, \cdots, S_h, (S_1 / \{a\}) \cup \{p\}, (S_1 / \{b\} ) \cup \{p\}.$ Note that from our assumptions on $p$, all of these $h+1$ sets are serene, and they all have value $\le m < h.$ We now consider two cases. Case 1. $p < \frac{h+2}{2}$ In this case, if there exists a serene set with value at least $p$ then we can finish by induction. Note that if we remove any serene set of size $m$, we know from induction that there are at least $(h-2)p + 1$ enmity relations among the remaining $(h-1)p$ people. Hence, the sum of all degrees is at least $2(h-2)p + 2m + 2.$ So if we return to our $h+1$ sets from earlier it's clear that one of them has value at least $\frac{2(h-2)p + 2m + 2}{h+1}.$ Since we know that this is at most $m$, we conclude that $m \ge p$. So we can indeed finish by induction by removing a serene set with value $\ge p$. Case 2. $p > \frac{h+2}{2}$ In this case, we seek to show that $m \ge h.$ Note that from removing a serene set of size $m$ and applying induction on the removing graph, there are at least $\frac{h(h-1)}{2} + m$ edges, and so the sum of all degrees is at least $h(h-1) + 2m.$ So by again considering our $h+1$ serene sets we've: $$m \ge \frac{h(h-1) + 2m}{h+1},$$ which easily yields $m \ge h.$ The induction is complete, and therefore so is the solution. $\square$