Let $m$ be a positive integer. Consider a $4m\times 4m$ array of square unit cells. Two different cells are [i]related[/i] to each other if they are in either the same row or in the same column. No cell is related to itself. Some cells are colored blue, such that every cell is related to at least two blue cells. Determine the minimum number of blue cells.
1. **Understanding the Problem:**
   We are given a \(4m \times 4m\) array of square unit cells. Two cells are related if they are in the same row or column. Each cell must be related to at least two blue cells. We need to determine the minimum number of blue cells required to satisfy this condition.

2. **Initial Setup:**
   The answer is \(6m\). This can be achieved by placing copies of the \(4 \times 4\) matrix 
   \[
   \begin{bmatrix}
   1 & 1 & 1 & 1 \\
   1 & 1 & 1 & 1 \\
   1 & 1 & 1 & 1 \\
   1 & 1 & 1 & 1
   \end{bmatrix}
   \]
   down the diagonal of the \(4m \times 4m\) array.

3. **Contradiction Argument:**
   Suppose there exists a valid array with fewer than \(6m\) blue cells. 

4. **Column Analysis:**
   If there is a column with no blue cells, then each row must contain at least 2 blue cells. This would yield at least \(8m\) blue cells in the array, which is a contradiction. Hence, each column contains at least one blue cell.

5. **Classification of Blue Cells:**
   Define a blue cell as:
   - **Single**: if it belongs to a row with exactly 1 blue cell.
   - **Double**: if it belongs to a row with exactly 2 blue cells.
   - **Multi**: if it belongs to a row with 3 or more blue cells.

   Let \(a\), \(b\), and \(c\) denote the number of rows with singles, doubles, and multis, respectively. Thus, we have:
   \[
   a + b + c = 4m
   \]
   and
   \[
   a + 2b + 3c < 6m \quad (\star)
   \]

6. **Redundancy Calculation:**
   For a column with \(n\) blue cells, define the redundancy of the column as \(n - 1\). Let \(\mathcal{R}\) be the total redundancy of the array. We have:
   \[
   \mathcal{R} < 6m - 4m = 2m
   \]

7. **Redundancy Bound:**
   For a column \(\mathcal{C}\) with \(\alpha\) singles, \(\beta\) doubles, and \(\gamma\) multis, the redundancy of \(\mathcal{C}\) is at least \(\frac{2\alpha}{3} + \frac{\beta}{2}\). We need to show:
   \[
   \frac{\alpha}{3} + \frac{\beta}{2} + \gamma \ge 1
   \]
   If \(\gamma \ge 1\), the inequality holds trivially. If \(\gamma = 0\), then \(\mathcal{C}\) must have at least two doubles or at least three blue cells. In either case, the inequality holds.

8. **Summing Over All Columns:**
   Summing over all columns, we get:
   \[
   \frac{2a}{3} + b \le \mathcal{R} < 2m
   \]
   Multiplying this inequality by 3 and summing with \((\star)\), we obtain:
   \[
   3a + 5b + 3c < 12m
   \]
   But since \(a + b + c = 4m\), this yields a contradiction.

Thus, the minimum number of blue cells required is indeed \(6m\).

The final answer is \(\boxed{6m}\).