, 2023

Solution:
First, suppose that $x$ and $y$ were of different signs. Then $x y < 0$ and $x / y < 0$, but the set has at most one negative value, a contradiction. Hence, $x$ and $y$ have the same sign; without loss of generality, we say $x$ and $y$ are both positive.

Let $(s, d) := (x+y, x-y)$. Then the set given is equal to $\{s^2, d^2, \frac{1}{4}(s^2 - d^2), \frac{s+d}{s-d}\}$. We split into two cases:

- Case 1: $\frac{s+d}{s-d} = N$. This forces $s^2 = 28.8$ and $d^2 = 12.8$, since $\frac{1}{4}(28.8 - 12.8) = 4$. Then $s = 12 \sqrt{0.2}$ and $d = \pm 8 \sqrt{0.2}$, so $N$ is either $\frac{12+8}{12-8} = 5$ or $\frac{12-8}{12+8} = 0.2$.

- Case 2: $\frac{s+d}{s-d} \neq N$. Suppose $\frac{s+d}{s-d} = k$, so $(s, d) = ((k+1)t, (k-1)t)$ for some $t$. Then $s^2 : d^2 : \frac{1}{4}(s^2 - d^2) = (k+1)^2 : (k-1)^2 : k$. Trying $k = 4, 12.8, 28.8$ reveals that only $k = 4$ is possible, since $28.8 : 12.8 = (4-1)^2 : 4$. This forces $N = s^2 = \frac{5^2}{4} \cdot 12.8 = 80$.

Hence, our final total is $5 + 0.2 + 80 = 85.2$.