The positive integer $ n$ has the following property:
if the three last digits of $n$ are removed, the number $\sqrt[3]{n}$ remains. 
Find $n$.
1. Let \( y \) be the integer consisting of the three last digits of \( n \). Hence, there is an integer \( x \) such that \( n = 1000x + y \), where \( x \geq 0 \) and \( 0 \leq y < 1000 \).
2. By removing the three last digits of \( n \) (which is \( y \)), the remaining integer is \( x \), which is equal to \( \sqrt[3]{n} \). Therefore, we have:
   \[
   x = \sqrt[3]{1000x + y}
   \]
3. Cubing both sides of the equation, we get:
   \[
   x^3 = 1000x + y
   \]
4. Rearranging the equation, we find:
   \[
   y = x^3 - 1000x
   \]
5. Given that \( 0 \leq y < 1000 \), we substitute \( y = x^3 - 1000x \) into this inequality:
   \[
   0 \leq x^3 - 1000x < 1000
   \]
6. We need to find the integer \( x \) that satisfies this inequality. Let's test some values of \( x \):
   - For \( x = 0 \):
     \[
     y = 0^3 - 1000 \cdot 0 = 0
     \]
     This satisfies \( 0 \leq y < 1000 \), but it does not provide a non-trivial solution.
   - For \( x = 10 \):
     \[
     y = 10^3 - 1000 \cdot 10 = 1000 - 10000 = -9000
     \]
     This does not satisfy \( 0 \leq y < 1000 \).
   - For \( x = 20 \):
     \[
     y = 20^3 - 1000 \cdot 20 = 8000 - 20000 = -12000
     \]
     This does not satisfy \( 0 \leq y < 1000 \).
   - For \( x = 30 \):
     \[
     y = 30^3 - 1000 \cdot 30 = 27000 - 30000 = -3000
     \]
     This does not satisfy \( 0 \leq y < 1000 \).
   - For \( x = 31 \):
     \[
     y = 31^3 - 1000 \cdot 31 = 29791 - 31000 = -1209
     \]
     This does not satisfy \( 0 \leq y < 1000 \).
   - For \( x = 32 \):
     \[
     y = 32^3 - 1000 \cdot 32 = 32768 - 32000 = 768
     \]
     This satisfies \( 0 \leq y < 1000 \).

7. Therefore, \( x = 32 \) and \( y = 768 \) is a valid solution. Substituting these values back into the expression for \( n \):
   \[
   n = 1000x + y = 1000 \cdot 32 + 768 = 32768
   \]

The final answer is \( \boxed{32768} \).