Tadeo draws the rectangle with the largest perimeter that can be divided into $2015$ squares of sidelength $1$ $cm$ and the rectangle with the smallest perimeter that can be divided into $2015$ squares of sidelength $1$ $cm$. What is the difference between the perimeters of the rectangles Tadeo drew?
1. **Understanding the problem**:
   - We need to find the difference between the perimeters of the largest and smallest rectangles that can be divided into 2015 squares of side length 1 cm.
   - The area of each rectangle is \(2015 \, \text{cm}^2\) since each square has an area of \(1 \, \text{cm}^2\).

2. **Finding the dimensions for the largest perimeter**:
   - To maximize the perimeter, we need to maximize the difference between the side lengths \(a\) and \(b\) of the rectangle.
   - Since \(ab = 2015\), the largest difference occurs when one side is as large as possible and the other side is as small as possible.
   - The factors of 2015 are \(1\) and \(2015\).
   - Thus, the dimensions for the largest perimeter are \(a = 2015\) and \(b = 1\).

3. **Calculating the largest perimeter**:
   \[
   \text{Perimeter} = 2(a + b) = 2(2015 + 1) = 2 \times 2016 = 4032 \, \text{cm}
   \]

4. **Finding the dimensions for the smallest perimeter**:
   - To minimize the perimeter, we need to minimize the sum of the side lengths \(a\) and \(b\).
   - The smallest sum occurs when \(a\) and \(b\) are as close as possible.
   - We need to find the factors of 2015 that are closest to each other.
   - The prime factorization of 2015 is \(2015 = 5 \times 13 \times 31\).
   - The closest factors are \(31\) and \(65\) (since \(31 \times 65 = 2015\)).

5. **Calculating the smallest perimeter**:
   \[
   \text{Perimeter} = 2(a + b) = 2(31 + 65) = 2 \times 96 = 192 \, \text{cm}
   \]

6. **Finding the difference between the perimeters**:
   \[
   \text{Difference} = 4032 - 192 = 3840 \, \text{cm}
   \]

The final answer is \(\boxed{3840}\)