Proof We provide an algorithm to distribute the competitors.
Denote the rooms $A$ and $B$. At some initial stage, we move one person at a time from one room to the other one. We achieve the target by going through several adjustments. In every step of the algorithm, let $A$ and $B$ be the sets of competitors in room $A$ and $B$. Let $C(A)$ and $C(B)$ be the largest size of clique in room $A$ and $B$ respectively.

Step 1: Let $M$ be the largest clique of all competitors, $|M| = 2m$.
Move all members of $M$ to room $A$, and the remained ones to room $B$.
Since $M$ is the largest clique of all competitors, we have $C(A) = |M| \ge C(B)$.

Step 2: If $C(A) > C(B)$, move one person from room $A$ to room $B$. (In view of $C(A) > C(B)$, we have $A \ne \emptyset$.)
After every operation is done, $C(A)$ decreases by $1$ while $C(B)$ increases by $1$ at most. These operations will not end until
$$C(A) \le C(B) \le C(A) + 1.$$
At that time, we also have $C(A) = |A| \ge m$. (Otherwise there are at least $m+1$ members of $M$ in room $B$, at most $m-1$ members in room $A$, then $C(B) - C(A) \ge (m+1) - (m-1) = 2$, it is impossible.)

Step 3: Denote $K = C(A)$. If $C(B) = K$, we are done; or else $C(B) = K + 1$. From the above discussion,
$$K = |A| = |A \cap M| \ge m, \quad |B \cap M| \le m.$$

Step 4: If there is a clique $C$ in room $B$ with $|C| = K + 1$ and a competitor $x \in B \cap M$ but $x \notin C$, then move $x$ to room $A$, we are done.
In fact, after the operation, there are $K+1$ members of $M$ in room $A$, so $C(A) = K+1$. Since $x \notin C$, whose removal does not reduce $C$, $C(B) = C$. Therefore, $C(A) = C(B) = K+1$.
If such competitor $x$ does not exist, then each largest clique in room $B$ contains $B \cap M$ as a subset. In this case, we do step 5.

Step 5: Choose any of the largest clique $C$ ($|C| = K+1$) in room $B$, move a member of $C \setminus M$ to room $A$. (In view of $|C| = K+1 > m \ge |B \cap M|$, we know $C \setminus M \ne \emptyset$.)
Since we only move one person at a time from room $B$ to room $A$, so $C(B)$ decreases by $1$ at most. At the end of this step, we have $C(B) = K$.
Now, there is a clique $A \cap M$ in $A$, $|A \cap M| = K$. So $C(A) \ge K$.
We prove $C(A) = K$ as follows.
Let $Q$ be any clique of room $A$. We only need to show $|Q| \le K$.
In fact, the members of room $A$ can be classified into two:
(1) Some members of $M$ in view of $M$ being a clique, are friends with all the members of $B \cap M$.
(2) The members move from room $B$ to room $A$ at step 5, they are friends of $B \cap M$.
So, every member of $Q$ and members of $B \cap M$ are friends. What is more, $Q$ and $B \cap M$ both are cliques, so is $Q \cup (B \cap M)$.
Since $M$ is the largest clique of all competitors,
$$|M| \ge |Q \cup (B \cap M)| = |Q| + |B \cap M|$$
$$= |Q| + |M| - |A \cap M|.$$
So $|Q| \le |A \cap M| = K$. Therefore, after these 5 steps, we have $C(A) = C(B) = K$.