Problem 1928

Let us form a complete graph on the vertex set $S$ of countries, and for each edge $AB$, assign it a nonnegative integer value $d(A, B)$. For each vertex $A$, let $d(A) = \sum_{B \neq A} d(A, B)$. For any two vertex sets $\mathcal{A}$ and $\mathcal{B}$, let $d(\mathcal{A}, \mathcal{B}) = \sum_{A \in \mathcal{A}, B \in \mathcal{B}} d(A, B)$. Let $T$ be a vertex with the minimum $d(T)$. We shall prove the following strengthened proposition:
If $\min_{A \neq T} d(A) \ge k$, then there exist $A$ and $B$ such that for all vertex sets $\mathcal{A}$ and $\mathcal{B}$ satisfying $A \in \mathcal{A}, B \in \mathcal{B}$ and $\mathcal{A} \cup \mathcal{B} = S$, we have $d(\mathcal{A}, \mathcal{B}) \ge k$.
We prove this strengthened proposition by mathematical induction. When $n=3$, it is easy to see that regardless of whether $k=0,1,2$, as long as the villain chooses two countries other than $T$, the goal can be achieved.
Now suppose the proposition holds for all values from 3 to $n-1$. Then when there are $n$ countries, choose two countries $A$ and $B$ other than $T$. Given $A$ and $B$, suppose the division of countries that minimizes $d(\mathcal{A}, \mathcal{B})$ is $\mathcal{A}$ and $\mathcal{B}$, and without loss of generality assume $T \in \mathcal{B}$. Suppose the villain has not yet achieved his goal at this point, which means $d(\mathcal{A}, \mathcal{B}) < k$. If at this point $\mathcal{A} = \{A\}$, then since $d(A) \ge k$, we would have $d(\mathcal{A}, \mathcal{B}) = d(A) \ge k$, a contradiction, so $|\mathcal{A}| \ge 2$.
Let us merge all the countries in $\mathcal{B}$ into one large country $C$, and in the new graph, for all $X \in \mathcal{A}$, let $d(X, C)$ equal $\sum_{Y \in \mathcal{B}} d(X, Y)$ from the old graph. Then after merging the countries:
(i) the total number of countries is at most $n-1$ (since $\{B, T\} \subset \mathcal{B}$);
(ii) $C$ has the minimum diplomatic relations (since we know $d(\mathcal{A}, \mathcal{B}) < k$);
(iii) $\min_{X \neq C} d(X) \ge k$.

Therefore, by the induction hypothesis, we can find two countries $U$ and $V$ such that the villain achieves his goal. Note that $U \neq C$, otherwise, as long as all countries join the same ally as $U$, we would have $d(U, C) = d(C) < k$, a contradiction. Similarly, $V \neq C$.
Suppose at this point the countries' distribution into allies is $U$ and $V$. Define $XY = X \cap Y$, then we have
$$d(U, V) = d(U\mathcal{A}, V\mathcal{A}) + d(U\mathcal{A}, V\mathcal{B}) + d(U\mathcal{B}, V\mathcal{A}) + d(U\mathcal{B}, V\mathcal{B}). \quad (1)$$
By the induction hypothesis, we know that if the countries are divided into $U\mathcal{A}$ and $S - U\mathcal{A}$, then
$$d(U\mathcal{A}, S - U\mathcal{A}) = d(U\mathcal{A}, U\mathcal{B}) + d(U\mathcal{A}, V\mathcal{A}) + d(U\mathcal{A}, V\mathcal{B}) \geq k. \quad (2)$$
Moreover, based on the fact that $S - V\mathcal{B}$ and $V\mathcal{B}$ is also one way for the countries to be divided into allies when $A$ and $B$ are chosen, by the minimality of $d(\mathcal{A}, \mathcal{B})$ we know
$$\begin{aligned} d(U\mathcal{A}, V\mathcal{B}) + d(V\mathcal{A}, V\mathcal{B}) + d(U\mathcal{B}, V\mathcal{B}) &= d(S - V\mathcal{B}, V\mathcal{B}) \geq d(\mathcal{A}, \mathcal{B}) \\ &= d(\mathcal{A}U, BU) + d(\mathcal{A}V, BU) + d(\mathcal{A}U, BV) + d(\mathcal{A}V, BV), \end{aligned}$$
that is
$$d(U\mathcal{B}, V\mathcal{B}) \geq d(U\mathcal{A}, U\mathcal{B}) + d(U\mathcal{B}, V\mathcal{A}). \quad (3)$$
Combining (1)-(3), we have
$$\begin{aligned} d(U, V) &= d(U\mathcal{A}, V\mathcal{A}) + d(U\mathcal{A}, V\mathcal{B}) + d(U\mathcal{B}, V\mathcal{A}) + d(U\mathcal{B}, V\mathcal{B}) \\ &\geq d(U\mathcal{A}, V\mathcal{A}) + d(U\mathcal{A}, V\mathcal{B}) + d(U\mathcal{B}, V\mathcal{A}) + \{d(U\mathcal{A}, U\mathcal{B}) + d(U\mathcal{B}, V\mathcal{A})\} \\ &= \{d(U\mathcal{A}, U\mathcal{B}) + d(U\mathcal{A}, V\mathcal{A}) + d(U\mathcal{A}, V\mathcal{B})\} + 2 \times d(U\mathcal{B}, V\mathcal{A}) \geq k, \end{aligned}$$
and thus the proposition is proved.