There are $2024$ cities in a country, every two of which are bidirectionally connected by exactly one of three modes of transportation - rail, air, or road. A tourist has arrived in this country and has the entire transportation scheme. He chooses a travel ticket for one of the modes of transportation and the city from which he starts his trip. He wants to visit as many cities as possible, but using only the ticket for the specified type of transportation. What is the largest $k$ for which the tourist will always be able to visit at least $k$ cities? During the route, he can return to the cities he has already visited.

[i]Proposed by Bogdan Rublov[/i]
1. **Understanding the Problem:**
   We have a complete graph \( K_{2024} \) where each edge is colored with one of three colors (representing rail, air, or road). The goal is to determine the largest number \( k \) such that the tourist can always visit at least \( k \) cities using only one type of transportation.

2. **Graph Coloring and Monochromatic Subgraphs:**
   The problem can be translated into finding the largest monochromatic connected subgraph in a complete graph with edges colored in three colors. We need to find the minimum possible size of the largest monochromatic connected subgraph.

3. **Claim:**
   Let \( K_{m,n} \) be a complete bipartite graph with parts of \( m \) and \( n \) vertices. If its edges are colored in two colors, then there exists a connected subgraph with edges of the same color with at least \( \frac{m+n}{2} \) vertices.

4. **Proof of the Claim:**
   - Let \( A \) and \( B \) be the partitions of the vertices, with \( |A| = m \) and \( |B| = n \). Assume the colors are white and black.
   - For any \( a \in A \), let \( d_w(a) \) denote the number of white edges incident with \( a \) and \( d_b(a) \) denote the number of black edges.
   - Suppose without loss of generality that \( \sum_{a \in A} d_w(a) \ge \sum_{a \in A} d_b(a) \). This implies:
     \[
     \sum_{a \in A} d_w(a) \ge \frac{mn}{2}
     \]
   - Delete the black edges. Let the connected components of the remaining graph consist of \( (m_1, n_1), (m_2, n_2), \ldots, (m_k, n_k) \) vertices.
   - Assume for contradiction that:
     \[
     m_i + n_i < \frac{m+n}{2} \quad \text{for all } i
     \]
   - We have:
     \[
     \sum_{i=1}^k m_i n_i \ge \frac{mn}{2}
     \]
   - However, under the conditions \( m_i + n_i < \frac{m+n}{2} \) and \( \sum_i m_i = m \), \( \sum_i n_i = n \), it holds that:
     \[
     \sum_{i=1}^k m_i n_i < \frac{mn}{2}
     \]
   - This is a contradiction, thus proving the claim. \(\blacksquare\)

5. **Application to the Original Problem:**
   - Let \( K \) be a complete graph of order \( 4n \), where \( n \in \mathbb{N} \).
   - Consider the largest possible monochromatic component, and let its color be white. Let \( V \) be its vertex set.
   - Assume \( |V| < 2n \). Let \( V' := V(K) \setminus V \).
   - Consider the induced bipartite graph on the parts \( V \) and \( V' \). There is no edge between \( V \) and \( V' \) colored in white, because we took the maximum component.
   - By the above claim, there is a connected monochrome subgraph with at least \( \frac{4n}{2} = 2n \) vertices, which contradicts the maximality of \( V \).

6. **Construction of Coloring:**
   - Group the vertices into 4 groups \( A_1, A_2, B_1, B_2 \), each with \( n \) vertices.
   - Color edges as follows:
     - White: between \( A_1 \) and \( A_2 \), and between \( B_1 \) and \( B_2 \).
     - Black: between \( A_1 \) and \( B_1 \), and between \( A_2 \) and \( B_2 \).
     - Red: between \( A_1 \) and \( B_2 \), and between \( A_2 \) and \( B_1 \).

The final answer is \( \boxed{1012} \).