158*. Using the digits from 1 to 9 once each, form the smallest nine-digit number that is divisible by 11.
$\triangle$ The sum of all digits of the desired number is equal to

$$
1+2+3+\ldots+9=45
$$

Let the sum of the digits in odd positions be denoted by $S_{1}$, and the sum of the digits in even positions by $S_{2}$. Then $S_{1}+S_{2}=45$.

The difference $S_{1}-S_{2}$, according to the divisibility rule by 11, is divisible by 11. Moreover, it is odd, since the sum $S_{1}+S_{2}$ is odd, and the sum and difference of two integers have the same parity. Therefore, this difference can take the values 11, $-11, 33$, and -33. Let's consider four cases.

1) Suppose $S_{1}-S_{2}=11$.

From the system of equations

$$
\left\{\begin{array}{l}
S_{1}+S_{2}=45 \\
S_{1}-S_{2}=11
\end{array}\right.
$$

we find that $S_{1}=28, S_{2}=17$.

2) Suppose $S_{1}-S_{2}=-11$.

Then

$$
\left\{\begin{array}{l}
S_{1}+S_{2}=45 \\
S_{1}-S_{2}=-11
\end{array}\right.
$$

from which $S_{1}=17, S_{2}=28$

3) Suppose $S_{1}-S_{2}=33$.

From the system of equations

$$
\left\{\begin{array}{l}
S_{1}+S_{2}=45 \\
S_{1}-S_{2}=33
\end{array}\right.
$$

we get that $S_{1}=39, S_{2}=6$. However, the equality $S_{2}=6$ is impossible, since if this sum consists even of 4 (and not 5) addends $a, b, c$, and $d$, then

$$
S_{2}=a+b+c+d \geq 1+2+3+4>6
$$

4) Suppose $S_{1}-S_{2}=-33$.

Similarly, we establish that this case is also impossible. Thus, for $S_{1}$ and $S_{2}$, only the values 28 and 17 (in some order) remain.

Let's try to start the desired nine-digit number with: 12345; since we need the smallest number divisible by 11. The four remaining digits we will denote in sequential order by $k, l, m$, and $n$. We get the number $12345 \mathrm{klmn}$.

If the sum $1+3+5=9$ is supplemented to 28 with the digits $l$ and $n$, then $l+n=28-9=19$, which is impossible.

If the sum $2+4=6$ is supplemented to 28 with the digits $k$ and $m$, then $k+m=22$, which is also impossible.

Therefore, it is impossible to start the desired number with the five-digit number 12345, since then neither the sum $S_{1}$ nor the sum $S_{2}$ can be equal to 28.

Let's try to start the desired number with the four-digit number 1234, and denote the number itself as $\overline{1234 a b c d e}$. Since $1+3=4, 2+4=6$, then

$$
a+c+e=28-4=24, \quad b+d=17-6=11
$$

or

$$
a+c+e=17-4=13, \quad b+d=28-6=22
$$

The second option is immediately ruled out. Only the first remains:

$$
\left\{\begin{array}{l}
a+c+e=24 \\
b+d=11
\end{array}\right.
$$

In this case, $a, b, c, d$, and $e$ take different values from the set $\{5 ; 6 ; 7 ; 8 ; 9\}$.

The digit $a$ should be as small as possible. But the value $a=5$, as we have seen, is impossible; the value $a=6$ is also impossible, since then $c+e=18$, from which the digits $c$ and $e$ are equal: $c=e=9$. Let's try $a=7$. In this case:

$$
\left\{\begin{array}{l}
c+e=17 \\
b+d=11
\end{array}\right.
$$

From the first equation $c=8, e=9$ (but not $c=9, e=8$), and from the second $b=5, d=6$. Answer: 123475869.