The six pairwise sums split into three complementary pairs, each pair adding up to the total $a + b + c + d$ of the set

So $x + y = 3s - 1030,$ where the common pair total $s$ is a sum of two of the four given numbers; make $s$ as large as possible

Let the set be $\{a, b, c, d\}$ with total $s = a + b + c + d.$ The six pairwise sums come in three complementary pairs: $$\begin{aligned} &(a+b) + (c+d) \\ &= (a+c) + (b+d) \\ &= (a+d) + (b+c) = s. \end{aligned}$$ No two of the pairings of $189,$ $320,$ $287,$ $234$ into two pairs give equal totals ($509 \ne 521,$ $476 \ne 554,$ $423 \ne 607$), so $x$ and $y$ are not paired with each other; each is paired with a given sum, and the remaining two given sums are paired together. Adding all six values, $x + y$ $= 3s - (189 + 320 + 287 + 234)$ $= 3s - 1030,$ where $s$ is the sum of two of the given numbers.

The largest choice is $s = 320 + 287 = 607,$ giving $x + y = 3 \cdot 607 - 1030 = 791.$ This is attained by the set $\{51.5,\ 137.5,\ 182.5,\ 235.5\},$ whose pairwise sums are $189,$ $234,$ $287,$ $320,$ $373,$ and $418,$ with $373 + 418 = 791.$