TITLE: 200 different natural numbers to split in sum and products
QUESTION [9 upvotes]: I wonder if there are $n=200$ different natural numbers with the property that the sum (products) of the first $100$ is equal accordingly to the sum (products) of the rest $100$ numbers? How to find the numbers?  There are numbers like $2+8+9=3+4+12=19$  and $2⋅8⋅9=3⋅4⋅12$, but what about  $n>6$?

REPLY [2 votes]: To complement @Thomas Andrews' answer, here is an $n=8$ solution:
$$ \{1,6,7,10\}\quad\text{and}\quad \{2,3,5,14\},$$
and the Matlab code I used to get it (there are many others):
r = 14;

for i1 = 1:r
    for i2 = i1+1:r
        for i3 = i2+1:r
            for i4 = i3+1:r
                for j1 = 1:r
                    for j2 = j1+1:r
                        for j3 = j2+1:r
                            for j4 = j3+1:r
                                if((i1 ~= j1) && (i1 ~= j2) && (i1 ~= j3) && (i1 ~= j4))
                                    if((i2 ~= j1) && (i2 ~= j2) && (i2 ~= j3) && (i2 ~= j4))
                                        if((i3 ~= j1) && (i3 ~= j2) && (i3 ~= j3) && (i3 ~= j4))
                                            if((i4 ~= j1) && (i4 ~= j2) && (i4 ~= j3) && (i4 ~= j4))
                                                if((i1+i2+i3+i4 == j1+j2+j3+j4) && (i1*i2*i3*i4 == j1*j2*j3*j4))
                                                    disp([i1,i2,i3,i4,j1,j2,j3,j4])
                                                end
                                            end
                                        end
                                    end
                                end
                            end
                        end
                    end
                end
            end
        end
    end
end