Determine all $n \in \mathbb{N}$ for which [list][*] $n$ is not the square of any integer, [*] $\lfloor \sqrt{n}\rfloor ^3$ divides $n^2$. [/list]
1. Suppose \( n = k^2 + c \) where \( 0 < c < 2k + 1 \). Then \( \lfloor \sqrt{n} \rfloor = k \).

2. We need to check the condition \( k^3 \mid n^2 \). Substituting \( n = k^2 + c \), we get:
   \[
   k^3 \mid (k^2 + c)^2
   \]
   Expanding the right-hand side:
   \[
   (k^2 + c)^2 = k^4 + 2k^2c + c^2
   \]
   Therefore, we need:
   \[
   k^3 \mid k^4 + 2k^2c + c^2
   \]

3. Suppose \( \gcd(k, c) = d \) and let \( k = d \cdot k_1 \) and \( c = d \cdot c_1 \). Then:
   \[
   k^3 \mid (k^2 + c)^2 \implies d^3 k_1^3 \mid d^2 (d k_1^2 + c_1)^2
   \]
   Simplifying the right-hand side:
   \[
   d^3 k_1^3 \mid d^2 (d^2 k_1^4 + 2d k_1^2 c_1 + c_1^2)
   \]
   Dividing both sides by \( d^3 \):
   \[
   k_1^3 \mid d k_1^4 + 2 c_1 k_1^2 + \frac{c_1^2}{d}
   \]

4. Since \( k_1^3 \mid d k_1^4 + 2 c_1 k_1^2 + \frac{c_1^2}{d} \), we need:
   \[
   k_1^2 \mid 2 c_1 k_1^2 + \frac{c_1^2}{d}
   \]
   This implies:
   \[
   k_1^2 \mid c_1^2
   \]
   Therefore, \( k_1 \mid c_1 \).

5. Given \( \gcd(k_1, c_1) = 1 \), the only possibility is \( k_1 = 1 \). Hence, \( d = k \).

6. Substituting \( k_1 = 1 \) and \( d = k \), we get:
   \[
   n = k^2 + c_1 k
   \]
   where \( c_1 = 1 \) or \( 2 \).

7. For \( c_1 = 1 \):
   \[
   n = k^2 + k
   \]
   We need \( k \mid k^2 + 2k + 1 \):
   \[
   k \mid k^2 + 2k + 1 \implies k \mid 1 \implies k = 1
   \]
   Therefore:
   \[
   n = 1^2 + 1 = 2
   \]

8. For \( c_1 = 2 \):
   \[
   n = k^2 + 2k
   \]
   We need \( k \mid k^2 + 4k + 4 \):
   \[
   k \mid k^2 + 4k + 4 \implies k \mid 4
   \]
   Therefore, \( k = 1, 2, \) or \( 4 \):
   \[
   \begin{cases}
   k = 1 & \implies n = 1^2 + 2 \cdot 1 = 3 \\
   k = 2 & \implies n = 2^2 + 2 \cdot 2 = 8 \\
   k = 4 & \implies n = 4^2 + 2 \cdot 4 = 24
   \end{cases}
   \]

Hence, the values of \( n \) that satisfy the given conditions are \( n = 2, 3, 8, 24 \).

The final answer is \( \boxed{2, 3, 8, 24} \).