How do you figure out what are all the possible numbers that four variables can be for a given N? I know what I want to do but I do not know how to do it. This is complication that I have, there are four variables A, B, C, D which all are ≥ 1. So in the equation (A+B+C+D)=N where I know what N equal to; how do I find the possible values of A, B, C, D?

EDIT:
To clarify A, B, C, D are four variables with the value of ≥ 1 and the relationship that they have the sum of one number.
So, in the equation... 
(A+B+C+D)=N
How do you figure out what are all the possible numbers that A, B, C, D can be for a given N?

P.S. I have just started learning mathematics with a new passion, I really enjoy it and have been relearning and learning a lot of stuff. If you could also point me in the direction of information that work with problems like these, that would be great!  
 A: Here are two possible answers given two different interpretations of your question.  I'm guessing you're interested in the first answer.
Answer 1
If by variable you mean positive integers, and $N$ is also a positive integer, then you can think of a solution to the equation $A+B+C+D=N$ as a placement of $N$ balls into $4$ boxes with the requirement that each box has at least $1$ ball.  This is the same as placing $N-4$ balls into four boxes with no requirements.  The answer to the number of ways to do this is
$$\binom{N-1}{3}=\frac{(N-1)!}{3!(N-4)!}=\frac{(N-1)(N-2)(N-3)}{6}=\frac{1}{6}N^3-N^2+\frac{11}{6}N-1$$
Notice this only makes sense if $N\ge 4$.  To see this, consider a collection of symbols consisting of $N-4$ circles (representing the balls) and $3$ vertical lines (representing a separation of the balls).  Altogether, there are $N-1$ symbols.  Orderings of these symbols correspond exactly to placements of the balls into four boxes, where we treat the vertical lines as the markers for when one box stops and another begins (notice we only need $3$ of them to delineate $4$ boxes).  To order these symbols, we really only need to choose where we put the $3$ lines, which gives the result above.
This isn't an algorithm for writing down all possible solutions, but it lets you know how many you need to write down.
Answer 2
If by variables you mean real numbers, then the equation $A+B+C+D=N$ with the requirement that $A,B,C$, and $D$ all be no less than $1$ has infinitely many solutions for $N>4$, 1 solution for $N=4$, and no solutions for $N< 4$.  I won't elaborate here unless this is actually the question you were asking.
A: To list them all, you're going to need to use some kind of algorithm.  The most obvious is three nested for loops.  For example, in GAP we have:
N:=8;;
for A in [1..N-3] do
  for B in [1..N-A-2] do
    for C in [1..N-A-B-1] do
      D:=N-A-B-C;
      Print([A,B,C,D],"\n");
    od;
  od;
od;

Above, some tweaks are made so that $A,B,C,D$ are all $\geq 1$ (if we didn't make these tweaks in the code above, $D$ might end up being $0$ or negative; this could be handled by not outputting the answer when $D \leq 0$).  I chose N:=8;; arbitrarily.
This will output:
[ 1, 1, 1, 5 ]
[ 1, 1, 2, 4 ]
[ 1, 1, 3, 3 ]
[ 1, 1, 4, 2 ]
[ 1, 1, 5, 1 ]
[ 1, 2, 1, 4 ]
[ 1, 2, 2, 3 ]
[ 1, 2, 3, 2 ]
[ 1, 2, 4, 1 ]
[ 1, 3, 1, 3 ]
[ 1, 3, 2, 2 ]
[ 1, 3, 3, 1 ]
[ 1, 4, 1, 2 ]
[ 1, 4, 2, 1 ]
[ 1, 5, 1, 1 ]
[ 2, 1, 1, 4 ]
[ 2, 1, 2, 3 ]
[ 2, 1, 3, 2 ]
[ 2, 1, 4, 1 ]
[ 2, 2, 1, 3 ]
[ 2, 2, 2, 2 ]
[ 2, 2, 3, 1 ]
[ 2, 3, 1, 2 ]
[ 2, 3, 2, 1 ]
[ 2, 4, 1, 1 ]
[ 3, 1, 1, 3 ]
[ 3, 1, 2, 2 ]
[ 3, 1, 3, 1 ]
[ 3, 2, 1, 2 ]
[ 3, 2, 2, 1 ]
[ 3, 3, 1, 1 ]
[ 4, 1, 1, 2 ]
[ 4, 1, 2, 1 ]
[ 4, 2, 1, 1 ]
[ 5, 1, 1, 1 ]

If you're after a mathematical description, the definition you give is already quite succinct; we could write it in set notation as: $$\{(A,B,C,D) \in (\mathbb{Z}_{\geq 1})^4:A+B+C+D=N\}.$$
