The question has provided $$\begin{align} a+b+c+d&=402\\ a+8&=d\\ c+4&=b \end{align}$$ and that the largest and smallest integers are 2 less than the sum of the other two integers.
From these, I have managed to conclude that there are $6$ different pairs I need to find but I've only managed to find $4$.
$$\begin{align} b+d&=207\\ a+c&=195\\ a+b&=199\\ c+d&=203 \end{align}$$
Am I going about things the right way? The end goal is to find what $a$,$b$,$c$, and $d$ are so I've done things in this way but I honestly can't figure out a way to figure out the last two pairs, which would be $a+d$ and $b+c$.
The largest plus smallest is $200$ and the other two must add to $202$.
From the results you have already obtained,the largest and smallest must be $a$ and $d$. Then $d=a+8$ forces $a=96,d=104$. Then $b=103,c=99$.