Consider the $4\times4$ array of $16$ dots, shown below.

[asy]
size(2cm);
dot((0,0));
dot((1,0));
dot((2,0));
dot((3,0));
dot((0,1));
dot((1,1));
dot((2,1));
dot((3,1));
dot((0,2));
dot((1,2));
dot((2,2));
dot((3,2));
dot((0,3));
dot((1,3));
dot((2,3));
dot((3,3));
[/asy]

Counting the number of squares whose vertices are among the $16$ dots and whose sides are parallel to the sides of the grid, we find that there are nine $1\times1$ squares, four $2\times2$ squares, and one $3\times3$ square, for a total of $14$ squares. We delete a number of these dots. What is the minimum number of dots that must be deleted so that each of the $14$ squares is missing at least one vertex?
To solve this problem, we need to ensure that each of the 14 squares (nine \(1 \times 1\) squares, four \(2 \times 2\) squares, and one \(3 \times 3\) square) is missing at least one vertex by deleting the minimum number of dots. 

1. **Identify the vertices of each square:**
   - There are 16 dots arranged in a \(4 \times 4\) grid.
   - Each \(1 \times 1\) square has 4 vertices.
   - Each \(2 \times 2\) square has 4 vertices.
   - The \(3 \times 3\) square has 4 vertices.

2. **Determine the overlap of vertices among the squares:**
   - The \(1 \times 1\) squares overlap at the center points of the grid.
   - The \(2 \times 2\) squares also overlap at the center points.
   - The \(3 \times 3\) square does not overlap with the center points but includes the corner points.

3. **Strategize the deletion of dots:**
   - To minimize the number of deletions, we should focus on deleting dots that are common to the greatest number of squares.
   - The center points of the grid are shared by multiple \(1 \times 1\) and \(2 \times 2\) squares.

4. **Delete the center points:**
   - The center points are \((1,1)\), \((1,2)\), \((2,1)\), and \((2,2)\).
   - Deleting these 4 center points will ensure that all \(1 \times 1\) and \(2 \times 2\) squares are missing at least one vertex.

5. **Check the \(3 \times 3\) square:**
   - The \(3 \times 3\) square does not include the center points but includes the corner points.
   - We need to delete one of the corner points to ensure the \(3 \times 3\) square is also missing a vertex.

6. **Adjust the deletions:**
   - If we delete one of the corner points, we need to ensure that the \(1 \times 1\) and \(2 \times 2\) squares still have at least one vertex missing.
   - By deleting one additional corner point, we can ensure that the \(3 \times 3\) square is missing a vertex without affecting the other squares.

7. **Final deletions:**
   - Delete the 3 center points: \((1,1)\), \((1,2)\), and \((2,1)\).
   - Delete one corner point: \((0,0)\).

Thus, we have deleted a total of 4 points.

The final answer is \(\boxed{4}\).