Determine the number of permutations $a_1, a_2, \dots, a_n$ of $1, 2, \dots, n$ such that for every positive integer $k$ with $1 \le k \le n$, there exists an integer $r$ with $0 \le r \le n - k$ which satisfies
\[ 1 + 2 + \dots + k = a_{r+1} + a_{r+2} + \dots + a_{r+k}. \]
1. **Define the problem in terms of permutations:**
   We need to determine the number of permutations \(a_1, a_2, \dots, a_n\) of \(1, 2, \dots, n\) such that for every positive integer \(k\) with \(1 \le k \le n\), there exists an integer \(r\) with \(0 \le r \le n - k\) which satisfies
   \[
   1 + 2 + \dots + k = a_{r+1} + a_{r+2} + \dots + a_{r+k}.
   \]

2. **Define \(n\)-cool permutations:**
   A permutation \((a_1, a_2, \dots, a_n)\) of \((1, 2, \dots, n)\) is called \(n\)-cool if it satisfies the given condition. Let \(S_n\) be the number of \(n\)-cool permutations.

3. **Consider the case \(k = n-1\):**
   For \(k = n-1\), the condition becomes:
   \[
   1 + 2 + \dots + (n-1) = a_1 + a_2 + \dots + a_{n-1} \quad \text{or} \quad 1 + 2 + \dots + (n-1) = a_2 + a_3 + \dots + a_n.
   \]
   This implies:
   \[
   \frac{(n-1)n}{2} = a_1 + a_2 + \dots + a_{n-1} \quad \text{or} \quad \frac{(n-1)n}{2} = a_2 + a_3 + \dots + a_n.
   \]
   Since the sum of the first \(n-1\) natural numbers is \(\frac{(n-1)n}{2}\), and the sum of all \(n\) numbers is \(\frac{n(n+1)}{2}\), it follows that:
   \[
   a_n = n \quad \text{or} \quad a_1 = n.
   \]

4. **Recursive relation:**
   If \(a_n = n\), then \((a_1, a_2, \dots, a_{n-1})\) must be an \((n-1)\)-cool permutation. Similarly, if \(a_1 = n\), then \((a_2, a_3, \dots, a_n)\) must be an \((n-1)\)-cool permutation. Therefore, any \((n-1)\)-cool permutation can be extended to an \(n\)-cool permutation by adding \(n\) to the front or the end of the permutation.

5. **Establish the recurrence relation:**
   Thus, we have:
   \[
   S_n = 2S_{n-1}.
   \]

6. **Base case:**
   For \(n = 1\), there is only one permutation \((1)\), which is trivially 1-cool. Hence:
   \[
   S_1 = 1.
   \]

7. **Solve the recurrence relation:**
   Using the recurrence relation \(S_n = 2S_{n-1}\) and the base case \(S_1 = 1\), we can solve for \(S_n\):
   \[
   S_2 = 2S_1 = 2 \cdot 1 = 2,
   \]
   \[
   S_3 = 2S_2 = 2 \cdot 2 = 4,
   \]
   \[
   S_4 = 2S_3 = 2 \cdot 4 = 8,
   \]
   and so on. By induction, we find:
   \[
   S_n = 2^{n-1}.
   \]

The final answer is \(\boxed{2^{n-1}}\).