1. Given that $x$ is a four-digit number, and the sum of its digits is $y$. When the value of $\frac{x}{y}$ is the smallest, $x=$ $\qquad$
$-1.1099$.
Let $x=\overline{a_{1} a_{2} a_{3} a_{4}}\left(a_{1} \in \mathbf{N}_{+}, a_{2} 、 a_{3} 、 a_{4} \in \mathbf{N}\right)$.
$$
\begin{array}{l}
\text { Then } \frac{x}{y}=\frac{1000 a_{1}+100 a_{2}+10 a_{3}+a_{4}}{a_{1}+a_{2}+a_{3}+a_{4}} \\
=1+9 \times \frac{111 a_{1}+11 a_{2}+a_{3}}{a_{1}+a_{2}+a_{3}+a_{4}} .
\end{array}
$$

To make $\frac{x}{y}$ the smallest, then $a_{4}=9$, at this time,
$$
\begin{array}{l}
\frac{x}{y}=1+9 \times \frac{111 a_{1}+11 a_{2}+a_{3}}{a_{1}+a_{2}+a_{3}+9} \\
=10+9 \times \frac{110 a_{1}+10 a_{2}-9}{a_{1}+a_{2}+a_{3}+9}
\end{array}
$$

To make $\frac{x}{y}$ the smallest, then $a_{3}=9$, at this time,
$$
\begin{array}{l}
\frac{x}{y}=10+9 \times \frac{110 a_{1}+10 a_{2}-9}{a_{1}+a_{2}+18} \\
=100+9 \times \frac{100 a_{1}-189}{a_{1}+a_{2}+18} .
\end{array}
$$

If $a_{1}=1$, then $\frac{x}{y}=100-9 \times \frac{89}{a_{2}+19}$.
To make $\frac{x}{y}$ the smallest, then $a_{2}=0$, at this time,
$$
x=1099, \frac{x}{y}=\frac{1099}{19} \text {. }
$$

If $a_{1}>1$, to make $\frac{x}{y}$ the smallest, then $a_{2}=9$, at this time,
$$
\begin{array}{l}
\frac{x}{y}=100+9 \times \frac{100 a_{1}-189}{a_{1}+27} \\
=1000-9 \times \frac{9 \times 3 \times 107}{a_{1}+27} .
\end{array}
$$

To make $\frac{x}{y}$ the smallest, then $a_{1}=2$, at this time,
$$
x=2999, \frac{x}{y}=\frac{2999}{29} \text {. }
$$

Since $\frac{2999}{29}>\frac{1099}{19}$, therefore, $x=1099$.