To solve this problem, we need to show that there exist two diametrically opposite cities on planet Zork where the prices of Kriptonita differ by at most 100 Urghs. We will use the properties of the roads and the cities given in the problem.

1. Graph Representation:
- Represent the cities on planet Zork as vertices of a graph $G$.
- Represent the roads between the cities as edges of the graph $G$.
- Since for every city $P$ there is a diametrically opposite city $P'$, we can consider an automorphism $\phi$ of order 2 on the graph $G$ such that $\phi(P) = P'$.

2. Properties of the Graph:
- The graph $G$ is connected, meaning there is a path between any two cities $P$ and $Q$.
- If there is a road between cities $P$ and $Q$, then there is also a road between $P'$ and $Q'$.
- The roads do not cross each other.

3. Price Difference Constraint:
- The prices of Kriptonita in two towns connected by a road differ by at most 100 Urghs.

4. Constructing the Argument:
- Consider the function $f: V(G) \to \mathbb{R}$ where $f(P)$ represents the price of Kriptonita in city $P$.
- For any edge $(P, Q) \in E(G)$, we have $|f(P) - f(Q)| \leq 100$.

5. Using the Automorphism:
- Since $\phi$ is an automorphism of order 2, it maps each city $P$ to its diametrically opposite city $P'$.
- We need to show that there exist cities $P$ and $P'$ such that $|f(P) - f(P')| \leq 100$.

6. Proof by Contradiction:
- Assume for contradiction that for all cities $P$, $|f(P) - f(P')| > 100$.
- Consider a path $P = P_0, P_1, P_2, \ldots, P_n = P'$ from $P$ to $P'$ in the graph $G$ (such a path exists because $G$ is connected).
- By the price difference constraint, we have:
$$|f(P_i) - f(P_{i+1})| \leq 100 \quad \text{for all } i = 0, 1, 2, \ldots, n-1.$$
- Summing these inequalities along the path, we get:
$$|f(P) - f(P')| \leq 100 \times n.$$
- Since $|f(P) - f(P')| > 100$ by assumption, this leads to a contradiction if $n = 1$ (i.e., $P$ and $P'$ are directly connected by a road).

7. Conclusion:
- Therefore, our assumption must be false, and there must exist at least one pair of diametrically opposite cities $P$ and $P'$ such that $|f(P) - f(P')| \leq 100$.

$\blacksquare$