1. Does there exist integers $a<b<c<d$ satisfying
$$
\frac{a}{a}+\frac{a}{b}+\frac{a}{c}+\frac{a}{d}=\frac{b}{a}+\frac{b}{b}+\frac{b}{c}+\frac{b}{d} ?
$$
1. Exist, one set of solutions is
$$
a=-28, b=-14, c=-7, d=4 \text {. }
$$

Obviously, it is impossible for all numbers to be positive (or all negative).

Assume $a=-4, b=-2, c=-1$, and solve for the corresponding $d$ that satisfies the requirements.
Substituting into the equation, we get
$$
\begin{array}{l}
\frac{-4}{-4}+\frac{-4}{-2}+\frac{-4}{-1}+\frac{-4}{d}=\frac{-2}{-4}+\frac{-2}{-2}+\frac{-2}{-1}+\frac{-2}{d} \\
\Rightarrow 1+2+4-\frac{4}{d}=\frac{1}{2}+1+2-\frac{2}{d} \\
\Rightarrow \frac{2}{d}=\frac{7}{2} \\
\Rightarrow d=\frac{4}{7} .
\end{array}
$$

Notice that the original equation is homogeneous. Therefore, all numbers can be multiplied by 7 simultaneously to make all solutions integers.