Determine if there are $2018$ different positive integers such that the sum of their squares is a perfect cube and the sum of their cubes is a perfect square.
1. Let \( c = 2018 \). We need to find \( 2018 \) different positive integers such that the sum of their squares is a perfect cube and the sum of their cubes is a perfect square.
2. Define \( A = \sum_{j=1}^{c} j^2 \) and \( B = \sum_{j=1}^{c} j^3 \).
3. Consider the sequence \( x_k = kA^4 B^3 \) for each \( k \in \mathbb{N} \).

4. Calculate the sum of the squares of \( x_k \):
   \[
   \sum_{k=1}^{c} x_k^2 = \sum_{k=1}^{c} (kA^4 B^3)^2 = A^8 B^6 \sum_{k=1}^{c} k^2
   \]
   Since \( \sum_{k=1}^{c} k^2 = A \), we have:
   \[
   \sum_{k=1}^{c} x_k^2 = A^8 B^6 A = A^9 B^6
   \]
   We need this to be a perfect cube. Notice that:
   \[
   A^9 B^6 = (A^3 B^2)^3
   \]
   Hence, \( \sum_{k=1}^{c} x_k^2 \) is indeed a perfect cube.

5. Calculate the sum of the cubes of \( x_k \):
   \[
   \sum_{k=1}^{c} x_k^3 = \sum_{k=1}^{c} (kA^4 B^3)^3 = A^{12} B^9 \sum_{k=1}^{c} k^3
   \]
   Since \( \sum_{k=1}^{c} k^3 = B \), we have:
   \[
   \sum_{k=1}^{c} x_k^3 = A^{12} B^9 B = A^{12} B^{10}
   \]
   We need this to be a perfect square. Notice that:
   \[
   A^{12} B^{10} = (A^6 B^5)^2
   \]
   Hence, \( \sum_{k=1}^{c} x_k^3 \) is indeed a perfect square.

6. Therefore, there exist \( 2018 \) different positive integers \( x_k = kA^4 B^3 \) that satisfy the conditions of the problem.

The final answer is \( \boxed{\text{True}} \)