Let $n$ be a given positive integer. Say that a set $K$ of points with integer coordinates in the plane is connected if for every pair of points $R, S\in K$, there exists a positive integer $\ell$ and a sequence $R=T_0,T_1, T_2,\ldots ,T_{\ell}=S$ of points in $K$, where each $T_i$ is distance $1$ away from $T_{i+1}$. For such a set $K$, we define the set of vectors 
\[\Delta(K)=\{\overrightarrow{RS}\mid R, S\in K\}\]
What is the maximum value of $|\Delta(K)|$ over all connected sets $K$ of $2n+1$ points with integer coordinates in the plane?

[i]Grigory Chelnokov, Russia[/i]
1. **Define the problem and the set \( K \)**:
   We are given a set \( K \) of \( 2n+1 \) points with integer coordinates in the plane. The set \( K \) is connected, meaning for any two points \( R, S \in K \), there exists a sequence of points \( R = T_0, T_1, \ldots, T_\ell = S \) in \( K \) such that each \( T_i \) is distance 1 away from \( T_{i+1} \).

2. **Define the set of vectors \( \Delta(K) \)**:
   \[
   \Delta(K) = \{ \overrightarrow{RS} \mid R, S \in K \}
   \]
   We need to find the maximum value of \( |\Delta(K)| \) over all connected sets \( K \) of \( 2n+1 \) points.

3. **Consider the structure of \( K \)**:
   Since \( K \) is connected, we can think of it as a graph where each point is a vertex and each edge represents a unit distance between two points. The number of edges in this graph is at least \( 2n \) because the graph must be connected and have \( 2n+1 \) vertices.

4. **Count the vectors**:
   Each edge in the graph corresponds to a vector of length 1. There are two types of unit vectors: horizontal \((1,0)\) and vertical \((0,1)\). Let \( a \) be the number of horizontal unit segments and \( b \) be the number of vertical unit segments. Then:
   \[
   a + b = 2n
   \]

5. **Calculate the number of distinct vectors**:
   Each horizontal segment contributes a vector \((1,0)\) and each vertical segment contributes a vector \((0,1)\). However, there can be additional vectors formed by combining these segments. For example, two horizontal segments can form a vector \((2,0)\), and two vertical segments can form a vector \((0,2)\).

6. **Count the extra vectors**:
   The number of extra vectors formed by combining horizontal segments is \( \binom{a}{2} \), and the number of extra vectors formed by combining vertical segments is \( \binom{b}{2} \). Therefore, the total number of extra vectors is:
   \[
   \binom{a}{2} + \binom{b}{2}
   \]

7. **Apply Karamata's inequality**:
   By Karamata's inequality, we have:
   \[
   \binom{a}{2} + \binom{b}{2} \ge 2 \binom{n}{2} = n^2 - n
   \]
   This inequality holds because the sum of the binomial coefficients is maximized when \( a \) and \( b \) are as equal as possible.

8. **Consider vectors with negative \( y \)-coordinates**:
   By considering vectors with negative \( y \)-coordinates, we double the number of extra vectors:
   \[
   2(n^2 - n) = 2n^2 - 2n
   \]

9. **Calculate the total number of vectors**:
   The total number of vectors is the sum of the unit vectors and the extra vectors:
   \[
   (2n)(2n+1) - (2n^2 - 2n) = 4n^2 + 2n - 2n^2 + 2n = 2n^2 + 4n
   \]

10. **Construct a configuration**:
    For a configuration that achieves this maximum, consider the points \((0,0), (1,0), \ldots, (n,0), (n,1), \ldots, (n,n)\). This configuration has \( 2n+1 \) points and achieves the maximum number of vectors.

The final answer is \( \boxed{ 2n^2 + 4n } \)