There is a board with the shape of an equilateral triangle with side $n$ divided into triangular cells with the shape of equilateral triangles with side $ 1$ (the figure below shows the board when $n = 4$). Each and every triangular cell is colored either red or blue. What is the least number of cells that can be colored blue without two red cells sharing one side?
[img]https://cdn.artofproblemsolving.com/attachments/0/1/d1f034258966b319dc87297bdb311f134497b5.png[/img]
To solve this problem, we need to determine the least number of blue cells such that no two red cells share a side. We will use a systematic approach to find the minimum number of blue cells required.

1. **Understanding the Problem:**
   - We have an equilateral triangle of side length \( n \) divided into smaller equilateral triangles of side length 1.
   - Each cell (small triangle) is either red or blue.
   - No two red cells can share a side.

2. **Pattern Observation:**
   - For \( n = 1 \), the triangle has 1 cell. If it is red, it satisfies the condition.
   - For \( n = 2 \), the triangle has 4 cells. We can color two cells red and two cells blue such that no two red cells share a side.
   - For \( n = 3 \), the triangle has 9 cells. We can color five cells red and four cells blue such that no two red cells share a side.

3. **General Pattern:**
   - We observe that the cells can be colored in a checkerboard pattern to ensure no two red cells share a side.
   - For an equilateral triangle of side length \( n \), the total number of cells is \( n^2 \).

4. **Checkerboard Coloring:**
   - In a checkerboard pattern, each row alternates colors.
   - For a row of length \( 2m + 1 \), there are \( \lceil \frac{2m+1}{2} \rceil \) red cells and \( \lfloor \frac{2m+1}{2} \rfloor \) blue cells.

5. **Counting Red and Blue Cells:**
   - For an equilateral triangle of side length \( n \), the number of red cells is approximately half of the total cells.
   - The number of red cells is \( \lfloor \frac{n^2 - 1}{2} \rfloor + k \), where \( n \in \{2k-1, 2k\} \) and \( k \ge 1 \).
   - The number of blue cells is \( \lfloor \frac{n^2 - 1}{2} \rfloor + 1 - k \).

6. **Induction Proof:**
   - **Base Case:** For \( n = 1 \), there is 1 cell, and it can be red.
   - **Inductive Step:** Assume the formula holds for \( n = k \). For \( n = k+1 \), we add a new row of \( k+1 \) cells. By the checkerboard pattern, the new row will have \( \lceil \frac{k+1}{2} \rceil \) red cells and \( \lfloor \frac{k+1}{2} \rfloor \) blue cells. This maintains the balance and ensures no two red cells share a side.

7. **Conclusion:**
   - The least number of blue cells required is given by the formula \( \lfloor \frac{n^2 - 1}{2} \rfloor + 1 - k \).

The final answer is \( \boxed{ \lfloor \frac{n^2 - 1}{2} \rfloor + 1 - k } \).