Yes. The generalization is provided by modular arithmetic. The properties you are observing all come from the fact that taking the remainder modulo $n$ respects addition and multiplication, and this generalizes to any $n$. More generally in abstract algebra we study rings and their ideals for the same reasons.
The notion of evenness and oddness of functions is closely related, but it is somewhat hard to explain exactly why. The key point is that there is a certain group, the cyclic group $C_2$ of order $2$, which is behind both concepts. For now, note that the product of two even functions is even, the product of an even and odd function is odd, and the product of two odd functions is even, so even and odd functions under multiplication behave exactly the same way as even and odd numbers under addition.
There are also huge generalizations depending on exactly what you're looking at, so it's hard to give a complete list here. You mentioned chessboards; there is a more general construction here, but it is somewhat hard to explain and there are no good elementary references that I know of. Once you learn some modular arithmetic, here is the modular arithmetic explanation of the chessboard idea: you can assign integer coordinates $(x, y)$ to each square (for example the coordinate of the lower left corner), and then you partition them into black or white squares depending on whether $x + y$ is even or odd; that is, depending on the value of $x + y \bmod 2$. Then given two points $(a, b)$ and $(c, d)$ you can consider the difference $c + d - a - b \bmod 2$, and constraints on this difference translate to constraints on the movement of certain pieces. This idea can be used, for example, to prove that certain chessboards (with pieces cut out of them) cannot be tiled with $1 \times 2$ or $2 \times 1$ tiles because these tiles must cover both a white square and a black square. Of course there are generalizations with $2$ replaced by a larger modulus and larger tiles.
As for matrices and vectors, let's just say that there are a lot of things this could mean, and none of them are straightforward generalizations of the above concept.
Several simplifications are possible. First, consider numbers with the same number of digits at a given time, so the range should be 100-999 instead of 1-1000. You have already established the result from 1-99, and if you really want 1-1000 you could just notice that 1000 needs to be counted.
It looks like when you say difference, you mean signed difference: 23 doesn't count because the sum of the evens is one less than the sum of the odds.
Consider the case of $n$ digits with $n$ even ($n$ odd will be similar). Let $m=\frac n2$. You are interested in how many ways there are to get $m$ numbers in the range $0$ to $9$ to make a given sum. So you can define $N(m,p$) as the number of ways to have $m$ numbers in this range add up to $p$ for the odd places. There is some perturbation for the odd places as you don't allow a leading zero. In this case, $N(2,1)=1, N(2,2)=2, \ldots N(2,9)=9, N(2,10)=9 , \ldots N(2,18)=1$ Then define $M(m,p)$ similarly for the even places, $M(2,0)=1, M(2,1)=2, \ldots M(2,9)=10, M(2,10)=9 , \ldots M(2,18)=1$. Now you can find $\sum N(m,i)M(m,i-1)$ to get the total.
Best Answer
Yes, it is possible. Divide your sum by $S$ and you have
$$1=\sum_i \frac {x_i}S$$
which is an Egyptian fraction expression of $1$ where all the denominators have the same number of factors of $2$. This is known to be solvable with all denominators odd, but all known solutions have an odd number of terms. A survey paper is here. The section of interest is $9.5$. One example is:
$$1=\frac 13+\frac 15+\frac 17 + \frac 19+\frac 1{11}+\frac 1{15}+\frac 1{35}+\frac 1{45}+\frac 1{231}$$
where the denominators have least common multiple $3465$ so we can write:
$$3465=1155+693+495+385+315+231+99+77+15$$
with every term dividing the sum. Now if we add $3465$ to each side we have a solution with an even number of terms:
$$6930=3465+1155+693+495+385+315+231+99+77+15$$
Any Egyptian fraction decomposition of $1$ into fractions with odd denominators yields a solution to your problem. The sum will be twice the least common multiple of the denominators in the decomposition. The paper shows that there is such a decomposition for all odd numbers of terms $9$ or above. You can multiply any solution by any odd number to get another.
What is happening is we are converting the Egyptian fraction decomposition of $1$ with all denominators odd into one that looks like $$1=\frac 12+\frac12\left(\text{all other terms}\right)$$