Brazilian Mathematical Olympiad

Let $a_i \ge b_i$ be the prices of Kriptonita in the antipodal towns $A_i$ and $B_i$ respectively. Suppose that the prices differ by more than 100 Urghs at all antipodal towns. Thus $a_i - b_i > 100$ for each $i$.
In a sequence of roads connecting $A_0$ to $B_0$, we can find a road connecting $A_j$ and $B_k$ for some $j, k$. Hence there exists a road connecting the antipodes $B_j$ and $A_k$ of $A_j$ and $B_k$.
If $j = k$ there exists a road connecting $A_j$ and $B_j$, which implies $a_j - b_j \le 100$, contradiction.
If $j \ne k$ then
$$\left| \begin{array}{l} a_j - b_k \le 100 \\ a_k - b_j \le 100 \end{array} \right. \implies a_j + a_k - b_j - b_k \le 200$$
On the other hand, we have
$$\left| \begin{array}{l} a_j - b_j > 100 \\ a_k - b_k > 100 \end{array} \right. \implies a_j + a_k - b_j - b_k > 200,$$
which is a contradiction.