Representation of number as a sums and differences of natural numbers Lets consider all the combinations of:
$$1+2+3+4=10,\ \ 1+2+3-4=2,\ \ 1+2-3+4=4,\ \ 1+2-3-4=-4, $$
$$1-2+3+4=6,\ \ 1-2+3-4=-2,\ \ 1-2-3+4=0,\ \ 1-2-3-4=-8,$$
$$-1+2+3+4=8,\ \ -1+2+3-4=0,\ \ -1+2-3+4=2,\ \ -1+2-3-4=-6,$$
$$-1-2+3+4=4,\ \ -1-2+3-4=-4,\ \ -1-2-3+4=-2,\ \ -1-2-3-4=-10$$
or
we have one 10, two 2, etc.
Let now create a list of tuples with integer as a first element and sum of the oddity of minus signs in a representation as a second one.
So we will have
$${(10,1),(8,-1),(6,-1),(4,-1+1),(2,-1+1),(0,1+1),(-2,1-1),(-4,1-1),(-6,-1),(-8,-1),(-10,1)} $$
or
$${(10,1),(8,-1),(6,-1),(4,0),(2,0),(0,2),(-2,0),(-4,0),(-6,-1),(-8,-1),(-10,1)} $$
or after removing the tuples with 0 second element:
$${(10,1),(8,-1),(6,-1),(0,2),(-6,-1),(-8,-1),(-10,1)} $$
Trying to create a generating function for the general case of the last sequence for even counts (for example 6, 8, etc instead of 4 elements). For the odd ones obviously all the second elements will be equal to $0$ due to oddity.
Actually I am familiar with creation of generating functions for restricted partitions, etc. But for this particular case not sure how to deal with negative numbers?

I believe you just need
$$ \prod_{k=1}^n (q^k - q^{-k}).$$
For example, $$(q-q^{-1})(q^2-q^{-2})(q^3-q^{-3})(q^4-q^{-4}) = q^{10} - q^8 - q^6 + 2 - q^{-6} - q^{-8} + q^{-10}$$
which corresponds to your final list of ordered pairs.
