Let any such

satisfying the desired result be called

** ***good* and let all other

be called

*bad*. I claim that

is good iff

is even.

First, we show

odd is bad.

Assume for the sake of contradiction that there is an odd number

that is good. Let

.

Consider the

-tuple:

, where there are

two's and

one. Assuming that this does indeed become an n tuple of equal positive integers, let this

-tuple be

. Note that

Now, considering the reverse operation of the "sumop" or the "pomus"

in which

we note that each pomus decreases both numbers. But, this will act on the numbers in pairs. As a result, one of the

's won't decrease.

Contradiction.

We show that all

even works.

Let

. Consider any

-tuple

. Form the

-tuple

where

. Observe that we can perform a "sumop" on

to get

(what I mean is that we add

). So, we get all of the

's to have the same highest power of

dividing each of them (like

,

,

). Let this new

-tuple be

Let

be the largest odd divisor of

. Let

and

be positive integers such that:

. We perform a "sumop" on

and

. Since

we note that we have a monovariant. So, since the

can't decrease infinitely, we find that eventually they are equal. Hence, we have all of them equal eventually.