200 different natural numbers to split in sum and products 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  $n>6$?
 A: 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

A: If you can do it for $n=2m$ numbers, you can do it for $n=2(m+3)$ numbers. Write your answer for $n=2m$ as $a_1,\dots,a_m,b_1,\dots,b_m$.
Choose $k>\max(a_1,\dots,a_m,b_1,\dots b_m)$ and add $a_{m+1}=2k,a_{m+2}=8k, a_{m+3}=9k,$ and $b_{m+1}=3k,b_{m+2}=4k,b_{m+3}=12k.$
Basically, we are adding a scaled version of you $n=3$ answer, picking the scale so that the new numbers are greater than all the previous numbers.
More generally, a solution for $n=2m$ and one for $n'=2m'$ gives a solution for $n''=2(m+m').$

We get an answer for $n=8$ with:
$$(2,9,16,18),\\
(3,6,12,24)$$
So you can get a sequence for $n=200$ by adding $6$ new elements to this $32$ times. 
Since you can pick $k$ to be a power of $2$, you can actually ensure that your sequence is all numbers of the form $2^i3^j$ where $j\leq 2.$

Given solutions for $n=6$ and $n=8$, we can solve for any $n=6a+8b$, for integers $a,b\geq 0.$ This is all even $n\geq 6$ other than $n=10.$ 
If you can solve for $n=10$, then you have solutions for all even $n\geq 6.$
Here is an answer for $n=10:$
$$(3,4,36,64,72),\\
(2,9,24,48,96)$$
So we preserve our property that we can find such sequences for all even $n\geq 8$ such that the numbers are of the form $2^i3^j$ with $j\leq 2$.

Here is how I found the $n=8$ example. The same was used for $n=10.$
First, added $1$ to both sides of the $n=3$ example. Then I multiplied both by $2$, getting:
$$(2,4,16,18),\\(2,6,8,24)$$
Then I added $9$ to both sequences:
$$(2,4,9,16,18),\\(2,6,8,9,24)$$
Then I replaced the $2,8,9$ on the second sequence with $(3,4,12)$ giving:
$$(2,4,9,16,18),\\(3,4,6,12,24)$$
Then I removed the common value $4$ from these two, giving:
$$(2,9,16,18),\\(3,6,12,24)$$
