Let $abcd=P$. Then \[\frac{P}{a}+a=\frac{P}{b}+b=\frac{P}{c}+c=\frac{P}{d}+d \]Thus, $\frac{P}{a}+a=\frac{P}{b}+b \implies P=ab \implies cd=1$ or $a=b$ Now, after this, you can do the same for each pair of numbers and do casework.