Let $n$ be a positive integer. David has six $n\times n$ chessboards which he arranges in an $n\times n\times n$ cube. Two cells are "aligned" if they can be connected by a path of cells $a=c_1,\ c_2,\ \dots,\ c_m=b$ such that all consecutive cells in the path share a side, and the sides that the cell $c_i$ shares with its neighbors are on opposite sides of the square for $i=2,\ 3,\ \dots\ m-1$. 

Two towers attack each other if the cells they occupy are aligned. What is the maximum amount of towers he can place on the board such that no two towers attack each other?
1. **Understanding the Problem:**
   We need to place the maximum number of towers on an \( n \times n \times n \) cube such that no two towers attack each other. Two cells are "aligned" if they can be connected by a path of cells where each consecutive pair of cells shares a side, and the sides shared are on opposite sides of the square.

2. **Construction:**
   The proposed construction involves placing towers on three faces of the cube that share a vertex. Specifically, we place towers on the main diagonals of three \( \left\lfloor \frac{n}{2} \right\rfloor \times \left\lfloor \frac{n}{2} \right\rfloor \) grids on these faces. This ensures that no two towers are aligned.

3. **Bounding the Number of Towers:**
   To find the upper bound on the number of towers, we use a double counting argument:
   - Each square without a tower is attacked at most twice.
   - Each tower attacks \( 8n - 2 \) squares.

4. **Double Counting Argument:**
   Let \( R \) be the number of towers. The total number of squares in the cube is \( 6n^2 \). Each square without a tower is attacked at most twice, so the total number of attacks on empty squares is at most \( 2(6n^2 - R) \).

   Each tower attacks \( 8n - 2 \) squares, so the total number of attacks by all towers is \( R(8n - 2) \).

   Therefore, we have:
   \[
   2(6n^2 - R) \geq R(8n - 2)
   \]

5. **Solving the Inequality:**
   \[
   2(6n^2 - R) \geq R(8n - 2)
   \]
   \[
   12n^2 - 2R \geq 8nR - 2R
   \]
   \[
   12n^2 \geq 8nR
   \]
   \[
   R \leq \frac{12n^2}{8n}
   \]
   \[
   R \leq \frac{3n}{2}
   \]

6. **Conclusion:**
   The maximum number of towers that can be placed on the \( n \times n \times n \) cube such that no two towers attack each other is \( \left\lfloor \frac{3n}{2} \right\rfloor \).

The final answer is \( \boxed{ \left\lfloor \frac{3n}{2} \right\rfloor } \).