Let $k=2j$ be even, and choose $k$ or more distinct primes, whose product $P$ is the radical of $k$ pairwise coprime numbers to be chosen.

Case $P$ is odd: Then the $2j$ numbers to be chosen are odd, and the sum is even, so $P$ is even, contradiction.

Case $P$ is even: Then, being pairwise coprime, one of the $2j$ numbers is even and the rest are odd, so the sum is odd, so $P$ is odd, contradiction.

So no such solutions for $k$ even.

You can investigate representations of odd squarefree numbers as a sum of powers or special multiples of their factors. I do not know of any examples. I suspect that the subject is a curiosity with few or no ties to other aspects of number theory, yet.

Gerhard "So Go And Make Some" Paseman, 2015.11.10