What is the largest possible number of subsets of the set $\{1, 2, \dots , 2n+1\}$ such that the intersection of any two subsets consists of one or several consecutive integers?
1. **Define the Problem and Setup:**
   We are given a set \( S = \{1, 2, \ldots, 2n+1\} \) and need to find the largest possible number of subsets such that the intersection of any two subsets consists of one or several consecutive integers.

2. **Reformulate the Problem:**
   Let \( A_i \subseteq S \) for \( 1 \leq i \leq N \). Define \( B_i = [\min A_i, \max A_i] \cap S \), which means \( B_i \) is an interval in \( S \). If \( B_i = B_j \) for some \( i \neq j \), then \( A_i = A_j \) because \( A_i \cap A_j \) is an interval.

3. **Intersection Property:**
   From \( A_i \cap A_j \subseteq B_i \cap B_j \), it follows that \( |B_i \cap B_j| \geq |A_i \cap A_j| \geq 2 \). Therefore, the family \( \mathcal{B} \) of the subsets \( B_i \) also has the mentioned property.

4. **Bounding the Number of Intervals:**
   Consider the intersection of all \( B_i \)'s, \( \bigcap_{i=1}^N B_i \), which contains an interval of length at least 1. Let this interval be \( [a+1, a+2] \). The number of such intervals is at most:
   \[
   (a+1)(n-a-1) = \left( \frac{m}{2} \right)^2 - \left( a - \frac{m-2}{2} \right)^2
   \]
   Therefore, \( N \leq \left\lfloor \frac{m^2}{4} \right\rfloor \).

5. **Example of Maximal Family:**
   An example of such a maximal family is the family of all intervals containing the interval \( \left[ \left\lfloor \frac{m}{2} \right\rfloor, \left\lfloor \frac{m}{2} \right\rfloor + 1 \right] \). This shows that the bound is sharp.

6. **Alternative Solution Using Graph Theory:**
   Build the graph \( G = (V, E) \) with \( V = \{1, 2, \ldots, m\} \) and \( E = \{ \{\min A, \max A\} \mid A \in \mathcal{A} \} \). Notice that \( \{\min A, \max A\} = \{\min A', \max A'\} \) implies \( A = A' \) since \( A \cap A' \) is an interval for \( A \neq A' \). Thus, \( |E| = |\mathcal{A}| = N \).

7. **Application of Mantel's Theorem:**
   The graph \( G \) can contain no triangle \( \{1 \leq i < j < k \leq m\} \) because then there would exist two members of \( \mathcal{A} \) intersecting in one point only. By Mantel's theorem (a special case of Turán's theorem), \( N \leq \left\lfloor \frac{m^2}{4} \right\rfloor \).

8. **Graph Construction:**
   The only model is a quasi-balanced complete bipartite graph, with its shores being \( \{\min A \mid A \in \mathcal{A}\} \) and \( \{\max A \mid A \in \mathcal{A}\} \), such that \( \max_{A \in \mathcal{A}} \min A < \min_{A \in \mathcal{A}} \max A \), corresponding to the construction shown above.

9. **Generalization:**
   If we take \( 1 \leq k \leq m \) and ask that for any \( 1 \leq i < j \leq N \) we have \( |A_i \cap A_j| \geq k \), with \( A_i \cap A_j \) being an interval, it follows that:
   \[
   N \leq \left\lfloor \left( \frac{m - k + 2}{2} \right)^2 \right\rfloor
   \]
   An example of such a maximal family is the family of all intervals containing the interval \( \left[ \left\lfloor \frac{m - k}{2} \right\rfloor + 1, \left\lfloor \frac{m - k}{2} \right\rfloor + k \right] \), showing that the bound is sharp.

The final answer is \( \boxed{ \left\lfloor \frac{m^2}{4} \right\rfloor } \).