To solve the given problem, we need to find all positive perfect cubes that are not divisible by $10$ and have the property that when the last three digits are erased, the resulting number is also a perfect cube.

1. Understanding the Cube Condition: Let $n^3$ be a perfect cube such that $n^3 \equiv 0 \pmod{1000}$. This means the last three digits of $n^3$ are zero, making $n^3$ a multiple of $1000$. However, the problem requires that $n^3$ should not be divisible by 10; that is, neither should $n^3$ end in zero nor should it be divisible by 5.

2. Formulation: Let $n^3 = 1000k + m$, where $0 \leq m < 1000$ represents the remaining last three digits. Given that erasing the last three digits of $n^3$ results in a perfect cube, we have $k = j^3$ for some integer $j$.

3. Range of Interest: We need to check for values $n^3 < 10000$ since we are initially interested in cubes where erasing might yield smaller cubes.

4. Computations:
- For $n^3 = 1331$, we have $n = 11$, since $1331 = 11^3$. When the last three digits are erased, the number is $1$, which is $1^3$.
- For $n^3 = 1728$, we have $n = 12$, since $1728 = 12^3$. When the last three digits are erased, the number is $1$, which is $1^3$.

5. Verification:
- Check if these cubes satisfy the divisibility condition. Both $1331 \equiv 1 \pmod{10}$ and $1728 \equiv 8 \pmod{10}$ are not divisible by $10$.
- Check if the condition upon erasing the last three digits is a cube. We already verified that in each case the number becomes $1$, which is indeed a cube since $1 = 1^3$.

Therefore, the perfect cubes that satisfy the conditions are:
$$\boxed{1331 \text{ and } 1728}.$$