jmc

For such a set $\{a,b,c,d\},$ the six pairwise sums can be themselves paired up into three pairs which all have the same sum: $$\begin{array}{rl}a+b& \text{ with }c+d,\\ a+c& \text{ with }b+d,\\ a+d& \text{ with }b+c.\end{array}$$Thus, the sum of all six pairwise sums is $3S,$ where $S=a+b+c+d,$ and so in our case, $$x+y=3S-(189+320+287+234)=3S-1030.$$Therefore, we want to maximize $S.$

Because of the pairing of the six pairwise sums, $S$ must be the sum of two of the four given numbers $189,$ $320,$ $287,$ and $234,$ so the greatest possible value of $S$ is $320+287=607.$ Therefore, the greatest possible value of $x+y$ is $3(607)-1030=791.$ This value is achievable for the set $\{51.5,137.5,182.5,235.5\},$ which has pairwise sums $189,$ $320,$ $287,$ $234,$ $373,$ and $418.$ Therefore the answer is $\boxed{791}.$