There are 2000 cities in Graphland; some of them are connected by roads. 
For every city the number of roads going from it is counted. It is known that there are exactly two equal numbers among all the numbers obtained. What can be these numbers?
1. **Understanding the Problem:**
   We are given a graph with 2000 cities (vertices) and some roads (edges) connecting them. For each city, we count the number of roads (degree of the vertex). It is known that there are exactly two cities with the same number of roads. We need to determine what these numbers can be.

2. **Graph Theory Basics:**
   - A graph \( G \) with \( n \) vertices can have vertex degrees ranging from 0 to \( n-1 \).
   - The sum of all vertex degrees in a graph is twice the number of edges (Handshaking Lemma).

3. **Induction Hypothesis:**
   We will use induction on the number of vertices \( 2n \). We claim that the common value of the two equal degrees can only be \( n \) or \( n-1 \).

4. **Base Case:**
   For \( n = 1 \), we have a graph with 2 vertices. The possible degrees are:
   - Both vertices have degree 0 (no edge).
   - Both vertices have degree 1 (one edge connecting them).
   In both cases, the degrees are equal, and the common value is 0 or 1, which fits our hypothesis.

5. **Inductive Step:**
   Assume the hypothesis holds for a graph with \( 2(n-1) \) vertices. We need to show it holds for \( 2n \) vertices.

6. **Case Analysis:**
   - **Case 1: There is a vertex with degree \( 2n-1 \):**
     - This vertex is connected to all other \( 2n-1 \) vertices.
     - There must be a vertex with degree 1 (connected only to the vertex with degree \( 2n-1 \)).
     - Remove these two vertices. The remaining graph has \( 2(n-1) \) vertices, each with degree reduced by 1.
     - By the induction hypothesis, the duplicated degree in this smaller graph is \( n-1 \).
     - Therefore, the duplicated degree in the original graph is \( n \).

   - **Case 2: There is a vertex with degree 0:**
     - This vertex is not connected to any other vertex.
     - The complementary graph \( \overline{G} \) has a vertex with degree \( 2n-1 \).
     - By the previous case, the duplicated degree in \( \overline{G} \) is \( n \).
     - Therefore, the duplicated degree in the original graph \( G \) is \( n-1 \).

7. **Conclusion:**
   The common value of the two equal degrees can only be \( n \) or \( n-1 \).

The final answer is \( \boxed{ n } \) or \( n-1 \).