Let $ m,n\in \mathbb{N}^*$. Find the least $ n$ for which exists $ m$, such that rectangle $ (3m \plus{} 2)\times(4m \plus{} 3)$ can be covered with $ \dfrac{n(n \plus{} 1)}{2}$ squares, among which exist $ n$ squares of length $ 1$, $ n \minus{} 1$ of length $ 2$, $ ...$, $ 1$ square of length $ n$. For the found value of $ n$ give the example of covering.
1. **Calculate the total area of the squares:**

   The total area of the squares is given by the sum of the areas of squares of different sizes. Specifically, we have:
   - \( n \) squares of side length \( 1 \)
   - \( n-1 \) squares of side length \( 2 \)
   - \( n-2 \) squares of side length \( 3 \)
   - ...
   - \( 1 \) square of side length \( n \)

   The total area \( A \) can be calculated as:
   \[
   A = n \cdot 1^2 + (n-1) \cdot 2^2 + (n-2) \cdot 3^2 + \cdots + 1 \cdot n^2
   \]

2. **Simplify the expression for the total area:**

   We can use the formula for the sum of squares and the sum of cubes to simplify the expression:
   \[
   A = \sum_{k=1}^{n} k^2 \cdot (n-k+1)
   \]

   This can be rewritten using the sum of squares and sum of cubes:
   \[
   A = (n+1) \sum_{k=1}^{n} k^2 - \sum_{k=1}^{n} k^3
   \]

   Using the formulas:
   \[
   \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}
   \]
   \[
   \sum_{k=1}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2
   \]

   Substituting these into the expression for \( A \):
   \[
   A = (n+1) \cdot \frac{n(n+1)(2n+1)}{6} - \left( \frac{n(n+1)}{2} \right)^2
   \]

3. **Simplify further:**

   \[
   A = \frac{n(n+1)^2(2n+1)}{6} - \frac{n^2(n+1)^2}{4}
   \]

   Finding a common denominator:
   \[
   A = \frac{2n(n+1)^2(2n+1) - 3n^2(n+1)^2}{12}
   \]

   Simplifying the numerator:
   \[
   A = \frac{2n(n+1)^2(2n+1) - 3n^2(n+1)^2}{12} = \frac{n(n+1)^2(2(2n+1) - 3n)}{12} = \frac{n(n+1)^2(n+2)}{12}
   \]

4. **Equate the area to the area of the rectangle:**

   The area of the rectangle is given by:
   \[
   (3m + 2)(4m + 3)
   \]

   Equating the two areas:
   \[
   \frac{n(n+1)^2(n+2)}{12} = (3m + 2)(4m + 3)
   \]

5. **Solve for \( m \):**

   We need to solve the quadratic equation:
   \[
   12m^2 + 17m + 6 = \frac{n(n+1)^2(n+2)}{12}
   \]

   Rearranging:
   \[
   12m^2 + 17m + 6 - \frac{n(n+1)^2(n+2)}{12} = 0
   \]

   Solving for \( m \):
   \[
   m = \frac{-17 + \sqrt{289 - 48 \left(6 - \frac{n(n+1)^2(n+2)}{12}\right)}}{24}
   \]

   Simplifying further:
   \[
   m = \frac{-17 + \sqrt{1 + 4n(n+1)^2(n+2)}}{24}
   \]

6. **Find the smallest \( n \):**

   We need \( m \) to be a natural number, so we solve:
   \[
   m = \frac{n^2 + 2n - 8}{12}
   \]

   For \( m \) to be an integer, \( n^2 + 2n - 8 \) must be divisible by 12. Testing values, we find:
   \[
   n \equiv 8 \pmod{12}
   \]

   The smallest such \( n \) is 8. For \( n = 8 \):
   \[
   m = \frac{8^2 + 2 \cdot 8 - 8}{12} = 6
   \]

   Therefore, the smallest \( n \) is 8, and for \( n = 8 \), \( m = 6 \).

The final answer is \( \boxed{ n = 8 } \).