Maximizing crazy positive integers

Let set {X1,X2, . . . ,Xn} be such that. X1+X2+ . . . +Xn = X1 * X2 * . . . * Xn...
then maximum possible value of max{...} will be n.
The Sum = Product = 2 * n.
The set would be {1,1, . . . 1,2,n } with (n-2) 1's
( Note:-If order matters we can find by permutation. )
First all are 1's, since last two ones are covered by the 2, and 1's are added to each location by last n. Thus the sum is 2 n, and multiplication is also 2 n.

If we had 3 as a member, product > addition for a set of maximums.

I have not studied Number theory. Is there some thing like this there??

In our case the answer is .. 5. +/ * =10.
Other sets of five are {1,1,2,2,2}. {1,1,1,3,3}

Let's say that $a\leq b\leq c\leq d\leq e$ . Rewrite the equation so it becomes

$e=\frac{a+b+c+d}{abcd-1}$

To maximize $e$ , we would want to minimize $abcd$ and maximize $a+b+c+b$ . We know that the rate at which $abcd$ increases is higher than the rate at which $a+b+c+d$ increases, so we would want to look at the case where $abcd$ is as small as possible. Since $abcd$ can't be 1, we turn to when $abcd = 2$ , or $a=1, b=1, c=1, d=2$ and $e=5$ .

A=1, b=2, c=3, d=4, e=5 so e is value of max integer