Here's a solution with some motivation attached to it.

The physical RMM sheet says $|A_1B_1|+|A_2B_2|>|A_1B_2|+|A_2B_1|$ but the version posted on the RMM website says $A_1B_1+A_2B_2>A_1B_2+A_2B_1$. Weird, fortunately it doesn't really affect anything...

pi37 wrote: Why was this placed in $3$-dimensional space with distances? Seems like an unnecessary red herring. It is easier to formulate it this way, than with introducing arbitrary symmetric real-valued function on the set of vertices. I hope that no participant thought that the nature of Euclidean distance is important here. By the way, in the origin of this problem the function was indeed a distance (but not necessary Euclidean).

Fedor Petrov wrote: It is easier to formulate it this way, than with introducing arbitrary symmetric real-valued function on the set of vertices. I hope that no participant thought that the nature of Euclidean distance is important here. By the way, in the origin of this problem the function was indeed a distance (but not necessary Euclidean).

and it may seem the right way forward is to do something with the triangle inequality or some another property of distances to dismiss those cases. I'd say that a big part of the difficulty of this problem comes from the extra information. It's similar to those geometry problems where the cyclicity condition of a quadrilateral is useless and the solution is pure projective geometry.

@ all the above, It is certainly conceivable that the geometry is important here: indeed, in 2 dimensions, with any convex quadrilateral $ABCD,$ $AC + BD > AB + CD$. I personally guessed that the geometric structure would be fairly useless, but there is certainly is reason for someone else to suspect the opposite.

Sorry for that. Triangle inequality is definitely a priori useless, since nothing changes in our process if we increase all 'distances' by the same value. Other geometric restrictions satisfied by points in three-dimensional space look too complicated and not related explicitly to sums of distances, do not they?

This might be identical with the above monovariant solutions, but I'm posting this for reference. Suppose that we can perform infinitely many vaild transformations. Since the number of all possible $\mathcal{A}\mathcal{B}$ trees are finite, some tree must appear twice during the procedure. Call such tree $\mathcal{T}$. Note that every vaild move switches an edge $ab$ to $ab'$, where $a \in \mathcal{A}$ and $b , b' \in \mathcal{B}$. Thus, in the sequence of moves from $\mathcal{T}$ to $\mathcal{T}$, there must exist a "pivot point" $a \in \mathcal{A}$ and endpoints $b_1, b_2, \cdots, b_t \in \mathcal{B}$ such that the moves $ab_1 \rightarrow ab_2 \rightarrow ab_3 \cdots \rightarrow ab_t \rightarrow ab_1$ are all valid. Now let $c_i$ be the unique vertex adjacent to both $b_i$ and $b_{i+1}$. (We consider indices modulo $t$ here.) Then, for each $1 \le i \le t$, $$|ab_i|+|c_ib_{i+1}| > |ab_{i+1}|+|c_ib_i|.$$Summing all gives $$\sum_{i} |c_ib_{i+1}| > \sum_{i} |c_{i}b_{i}|.$$We claim that each side has the same (multi)set of edges. Indeed, if then, $\sum_{i} |c_ib_{i+1}| > \sum_{i} |c_{i}b_{i}|=\sum_{i} |c_ib_{i+1}|$, which is impossible. Hence it suffices to prove the claim. Let us first reformulate the claim: Claim wrote: Let $\mathcal{T}$ be a $\mathcal{A}\mathcal{B}$ tree, and suppose that $b_1, b_2, \cdots, b_t \in \mathcal{B}$ satisfies $d(b_i, b_{i+1})=2$, and let $c_i$ be the unique vertex adjacent to both $b_i$ and $b_{i+1}$. Then the set of edges $X=\{ b_ic_i \}$ and $Y=\{b_i c_{i+1}\}$ are equal. Let us induct on $t$, the number of edges. The base case is easy. Suppose that all edges of the form $c_ib_i$ and $c_ib_{i+1}$ are distinct. Then, these edges form a cycle $b_1c_1b_2c_2b_3 \cdots c_{t-1}b_tc_tb_1$, which is impossible. Thus some two edge coincides. Note that if $\{ v, b_i \} = \{ b_{j}, w \}$ is an edge, then $b_ib_j$ is an edge, which is impossible since $\mathcal{B}$ is independent. Hence the only case is that $c_i=c_{i+1}$ for some $i$. For such case, by the induction hypothesis on $b_1, b_2, \cdots, b_{i}, b_{i+2}, \cdots, b_t$, $X- \{ b_{i+1}c_{i+1}\}=Y- \{ b_{i+1}c_i \}$. Hence, combining with $c_i=c_{i+1}$, we deduce that $X=Y$. This concludes the proof. $\blacksquare$