Let $N$ be the number of integer sequences $a_1, a_2, \dots, a_{2^{16}-1}$ satisfying \[0 \le a_{2k + 1} \le a_k \le a_{2k + 2} \le 1\] for all $1 \le k \le 2^{15}-1$. Find the number of positive integer divisors of $N$.

[i]Proposed by Ankan Bhattacharya[/i]
1. **Understanding the Problem:**
   We need to find the number of integer sequences \(a_1, a_2, \dots, a_{2^{16}-1}\) that satisfy the conditions:
   \[
   0 \le a_{2k + 1} \le a_k \le a_{2k + 2} \le 1
   \]
   for all \(1 \le k \le 2^{15}-1\).

2. **Binary Tree Interpretation:**
   The conditions suggest a binary tree structure where each node \(a_k\) has children \(a_{2k+1}\) and \(a_{2k+2}\). The values of the nodes are either 0 or 1, and the sequence must satisfy the given inequalities.

3. **Recursive Definition:**
   Let \(x_i\) be the number of valid sequences of depth \(i\) with the root labeled 0, and \(y_i\) be the number of valid sequences of depth \(i\) with the root labeled 1. The depth of the tree is 16.

4. **Recurrence Relations:**
   The recurrence relations for \(x_i\) and \(y_i\) are:
   \[
   x_i = x_{i-1} \cdot (x_{i-1} + y_{i-1})
   \]
   \[
   y_i = y_{i-1} \cdot (x_{i-1} + y_{i-1})
   \]
   Adding these, we get:
   \[
   x_i + y_i = (x_{i-1} + y_{i-1})^2
   \]

5. **Initial Conditions:**
   At depth 0, there is only one node, and it can be either 0 or 1:
   \[
   x_0 = 1, \quad y_0 = 1
   \]
   Therefore:
   \[
   x_0 + y_0 = 2
   \]

6. **Calculating \(x_{16} + y_{16}\):**
   Using the recurrence relation:
   \[
   x_i + y_i = (x_{i-1} + y_{i-1})^2
   \]
   We start with:
   \[
   x_0 + y_0 = 2
   \]
   Then:
   \[
   x_1 + y_1 = 2^2 = 4
   \]
   \[
   x_2 + y_2 = 4^2 = 16
   \]
   \[
   x_3 + y_3 = 16^2 = 256
   \]
   Continuing this process up to \(i = 16\):
   \[
   x_{16} + y_{16} = 2^{2^{16}}
   \]

7. **Number of Positive Integer Divisors:**
   The number \(N = 2^{2^{16}}\) is a power of 2. The number of positive integer divisors of \(2^k\) is \(k + 1\). Here, \(k = 2^{16}\):
   \[
   \text{Number of divisors} = 2^{16} + 1
   \]

The final answer is \(\boxed{65537}\)