Some blue and red circular disks of identical size are packed together to form a triangle. The top level has one disk and each level has 1 more disk than the level above it. Each disk not at the bottom level touches two disks below it and its colour is blue if these two disks are of the same colour. Otherwise its colour is red.

Suppose the bottom level has 2048 disks of which 2014 are red. What is the colour of the disk at the top?
1. **Understanding the Problem:**
   - We have a triangular arrangement of disks.
   - The bottom level has 2048 disks, of which 2014 are red.
   - Each disk not at the bottom level touches two disks below it.
   - A disk is blue if the two disks below it are of the same color, otherwise it is red.
   - We need to determine the color of the top disk.

2. **Induction Hypothesis:**
   - Let's denote the number of disks at level \( k \) as \( 2^k \).
   - We will use induction to show that the parity (even or odd number of red disks) of each level determines the color of the top disk.

3. **Base Case:**
   - For the bottom level \( k = 11 \) (since \( 2^{11} = 2048 \)), we have 2014 red disks.
   - The number of blue disks at this level is \( 2048 - 2014 = 34 \).

4. **Inductive Step:**
   - Assume that for level \( k \), the parity of the number of red disks determines the parity of the number of red disks at level \( k-1 \).
   - Consider the transition from level \( k \) to level \( k-1 \):
     - Each disk at level \( k-1 \) is determined by two disks at level \( k \).
     - If both disks at level \( k \) are of the same color, the disk at level \( k-1 \) is blue.
     - If the two disks at level \( k \) are of different colors, the disk at level \( k-1 \) is red.

5. **Parity Analysis:**
   - The parity of the number of red disks at level \( k \) affects the parity of the number of red disks at level \( k-1 \).
   - Specifically, if the number of red disks at level \( k \) is even, the number of red disks at level \( k-1 \) will also be even.
   - If the number of red disks at level \( k \) is odd, the number of red disks at level \( k-1 \) will be odd.

6. **Applying Induction:**
   - At the bottom level \( k = 11 \), we have 2014 red disks, which is even.
   - By induction, the parity of the number of red disks remains even as we move up each level.
   - Therefore, at the top level (level 0), the number of red disks is even.

7. **Conclusion:**
   - Since the top level has only one disk, and the number of red disks is even, the top disk must be blue.

The final answer is \(\boxed{\text{blue}}\).