HMMT February

Solution:
Let $a' = \frac{\operatorname{lcm}(a, b, c, d)}{a}$. Define $b'$, $c'$, and $d'$ similarly. We have that $a'$, $b'$, $c'$, and $d'$ are pairwise distinct positive integers that satisfy
$$\frac{1}{a'} + \frac{1}{b'} = \frac{1}{c'} + \frac{1}{d'}$$
Let $T$ be the above quantity. We have
$$a + b = T \operatorname{lcm}(a, b, c, d)$$
so we try to maximize $T$. Note that since $\frac{1}{2} + \frac{1}{3} < \frac{1}{1}$, we cannot have any of $a'$, $b'$, $c'$, and $d'$ be $1$. At most one of them can be $2$, so at least one side of the equation must have both denominators at least $3$. Hence, the largest possible value of $T$ is
$$T = \frac{1}{3} + \frac{1}{4} = \frac{1}{2} + \frac{1}{12} = \frac{7}{12}$$
and the second largest possible value of $T$ is
$$T = \frac{1}{3} + \frac{1}{5} = \frac{1}{2} + \frac{1}{30} = \frac{8}{15}.$$
Taking $T = \frac{7}{12}$ and $\operatorname{lcm}(a, b, c, d) = 996 = 12 \cdot 83$, we get $a + b = 581$. Since the next best value of $T$ gives $\frac{8}{15} \cdot 1000 < 534 < 581$, this is optimal.