1. Define the problem and notation:
- We have three buildings $A$, $B$, and $C$ each with 2015 floors.
- Each floor has one window, and each window can be colored red (1), green (2), or blue (3).
- Residents on the $k$-th floor can see the colors of the windows on the $k$-th and $(k-1)$-th floors of the other two buildings.
- We need to color the windows such that each resident can see at least one window of each color.

2. **Analyze the possible configurations for the $k$-th floor:**
- Denote the colors of the windows on the $k$-th floor as $T_k = (a_k, b_k, c_k)$.
- We need to determine the possible values for $T_{k+1} = (a_{k+1}, b_{k+1}, c_{k+1})$.

3. **Case 1: $T_k$ contains a single color:**
- Assume $T_k = (1, 1, 1)$.
- Each pair $(a_{k+1}, b_{k+1})$, $(b_{k+1}, c_{k+1})$, $(a_{k+1}, c_{k+1})$ must contain both numbers 2 and 3, which is impossible.
- Hence, no solution for $T_{k+1}$ in this case.

4. **Case 2: $T_k$ contains 2 distinct colors:**
- Assume $T_k = (1, 1, 2)$.
- The possible solutions for $T_{k+1}$ are $(2, 3, 3)$ and $(3, 2, 3)$.

5. **Case 3: $T_k$ contains 3 distinct colors:**
- Assume $T_k = (1, 2, 3)$.
- The possible solutions for $T_{k+1}$ are $(2, 3, 1)$ and $(3, 1, 2)$.

6. Conclusion for valid configurations:
- For a valid configuration of $T_k$ (2 or 3 distinct colors), there are exactly 2 possible configurations for $T_{k+1}$.

7. Calculate the number of ways to color the windows:
- The number of ways to color the windows is $N = n_1 \cdot 2^{2014}$, where $n_1$ is the number of valid configurations for $T_1$.

8. **Count the valid configurations for $T_1$:**
- For 2 distinct colors:
- Use twice the color $x$ and once the color $y$, where $x, y \in \{1, 2, 3\}$ and $x \neq y$.
- There are 6 possible ordered pairs $(x, y)$ and for each $(x, y)$ there are 3 triplets $(x, x, y)$, $(x, y, x)$, $(y, x, x)$.
- Hence, there are $n_2 = 6 \cdot 3 = 18$ possible triplets $T_1$ with 2 distinct colors.
- For 3 distinct colors:
- There are $n_3 = 3! = 6$ possible triplets $T_1$.
- Therefore, $n_1 = n_2 + n_3 = 18 + 6 = 24$.

9. Final calculation:
- The number of ways to color the windows is $N = 24 \cdot 2^{2014}$.