Let $a^{\prime}=\frac{\operatorname{lcm}(a, b, c, d)}{a}$. Define $b^{\prime}, c^{\prime}$, and $d^{\prime}$ similarly. We have that $a^{\prime}, b^{\prime}, c^{\prime}$, and $d^{\prime}$ are pairwise distinct positive integers that satisfy $$\frac{1}{a^{\prime}}+\frac{1}{b^{\prime}}=\frac{1}{c^{\prime}}+\frac{1}{d^{\prime}}$$ 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 $a^{\prime}, b^{\prime}, c^{\prime}$, and $d^{\prime}$ 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$8 / 15 \cdot 1000<534<581$, this is optimal.