Let $ n$ be a positive integer and $ [ \ n ] = a.$ Find the largest integer $ n$ such that the following two conditions are satisfied:

$ (1)$ $ n$ is not a perfect square;

$ (2)$ $ a^{3}$ divides $ n^{2}$.
To solve the problem, we need to find the largest integer \( n \) such that:
1. \( n \) is not a perfect square.
2. \( a^3 \) divides \( n^2 \), where \( a = \lfloor \sqrt{n} \rfloor \).

Let's denote \( n = k^2 + r \) where \( 0 \leq r < 2k + 1 \). This implies \( \lfloor \sqrt{n} \rfloor = k \). Therefore, \( a = k \).

Given that \( a^3 \) divides \( n^2 \), we have:
\[ k^3 \mid (k^2 + r)^2 \]

We will consider different cases for \( r \):

### Case 1: \( r = 0 \)
If \( r = 0 \), then \( n = k^2 \). This means \( n \) is a perfect square, which contradicts the first condition. Therefore, this case is not valid.

### Case 2: \( r = k \)
If \( r = k \), then \( n = k^2 + k \). We need to check if \( k^3 \mid (k^2 + k)^2 \):
\[
(k^2 + k)^2 = k^4 + 2k^3 + k^2
\]
Since \( k^3 \mid k^4 \) and \( k^3 \mid 2k^3 \), we need to check if \( k^3 \mid k^2 \):
\[
k^3 \mid k^2 \implies k \mid 1 \implies k = 1
\]
Thus, \( n = 1^2 + 1 = 2 \). This is a valid solution.

### Case 3: \( r = 2k \)
If \( r = 2k \), then \( n = k^2 + 2k \). We need to check if \( k^3 \mid (k^2 + 2k)^2 \):
\[
(k^2 + 2k)^2 = k^4 + 4k^3 + 4k^2
\]
Since \( k^3 \mid k^4 \) and \( k^3 \mid 4k^3 \), we need to check if \( k^3 \mid 4k^2 \):
\[
k^3 \mid 4k^2 \implies k \mid 4
\]
Thus, \( k \) can be \( 1, 2, \) or \( 4 \):
- If \( k = 1 \), then \( n = 1^2 + 2 \cdot 1 = 3 \).
- If \( k = 2 \), then \( n = 2^2 + 2 \cdot 2 = 8 \).
- If \( k = 4 \), then \( n = 4^2 + 2 \cdot 4 = 24 \).

### Conclusion
The valid values of \( n \) are \( 2, 3, 8, \) and \( 24 \). Among these, the largest value is \( 24 \).

The final answer is \( \boxed{24} \).