The colonizers of a spherical planet have decided to build $N$ towns, each having area $1/1000$ of the total area of the planet. They also decided that any two points belonging to different towns will have different latitude and different longitude. What is the maximal value of $N$?
1. **Understanding the Problem:**
   We need to determine the maximum number of towns, \( N \), that can be built on a spherical planet such that each town occupies \( \frac{1}{1000} \) of the total surface area of the planet. Additionally, any two points belonging to different towns must have different latitudes and longitudes.

2. **Surface Area Calculation:**
   The total surface area \( A \) of a sphere with radius \( r \) is given by:
   \[
   A = 4 \pi r^2
   \]
   Each town occupies an area of:
   \[
   A_{\text{town}} = \frac{4 \pi r^2}{1000}
   \]

3. **Latitude and Longitude Constraints:**
   Suppose the maximum and minimum latitudes of a town are given by points \( \ell_M \) and \( \ell_m \). The width \( w \) of the latitudinal band occupied by the town is:
   \[
   w = |\ell_M - \ell_m|
   \]
   The surface area of this band is proportional to its width:
   \[
   A_{\text{band}} = 2 \pi r w
   \]

4. **Longitudinal Band Calculation:**
   If the most easterly and westerly points of the town subtend an angle \( \theta \) with the center of the sphere, the area of the longitudinal band is:
   \[
   A_{\text{longitudinal}} = r \theta w
   \]
   The town must fit within this area:
   \[
   r \theta w = \frac{4 \pi r^2}{1000}
   \]

5. **Summing Widths and Angles:**
   Let \( \{w_i\}_{i=1}^N \) be the widths of the latitudinal bands made by the towns, and let \( \{\theta_i\}_{i=1}^N \) be the angles representing the widths of the longitudinal bands of the towns. Then:
   \[
   \sum_{i=1}^N w_i \le 2r \quad \text{and} \quad \sum_{i=1}^N \theta_i \le 2\pi
   \]

6. **Area of Each Town:**
   The area of town \( i \) is at most:
   \[
   A_i = r \theta_i w_i = \frac{4 \pi r^2}{1000}
   \]

7. **Applying Cauchy-Schwarz Inequality:**
   Using the Cauchy-Schwarz inequality:
   \[
   \left( \sum_{i=1}^N w_i \right) \left( \sum_{i=1}^N \theta_i \right) \ge \left( \sum_{i=1}^N \sqrt{w_i \theta_i} \right)^2
   \]
   Substituting the constraints:
   \[
   2r \cdot 2\pi \ge \left( \sum_{i=1}^N \sqrt{w_i \theta_i} \right)^2
   \]
   Since \( w_i \theta_i = \frac{4 \pi r}{1000} \):
   \[
   \left( \sum_{i=1}^N \sqrt{\frac{4 \pi r}{1000}} \right)^2 = N \left( \sqrt{\frac{4 \pi r}{1000}} \right)^2 = N \cdot \frac{4 \pi r}{1000}
   \]

8. **Solving for \( N \):**
   \[
   4 \pi r^2 \ge N \cdot \frac{4 \pi r}{1000}
   \]
   Simplifying:
   \[
   4 \pi r^2 \ge \frac{4 \pi r N}{1000}
   \]
   \[
   1000 r \ge N
   \]
   \[
   N \le \sqrt{1000}
   \]
   Since \( \sqrt{1000} \approx 31.62 \), the maximum integer value for \( N \) is:
   \[
   \max(N) = 31
   \]

The final answer is \( \boxed{31} \).