$10 \cdot 10$ Given a four-digit number that satisfies the following conditions:
(1) If the units digit and the hundreds digit, as well as the tens digit and the thousands digit, are simultaneously swapped, then its value increases by 5940;
(2) When divided by 9, the remainder is 8.
Find the smallest odd number among these four-digit numbers.
(Shandong Province, China Mathematics Competition, 1979)
［Solution］Since the number increased by 5940 after the positions were swapped, and its unit digit is 0, the unit digit and the hundredth digit of this four-digit number are the same.
Let this four-digit number be $\overline{a b c b}=1000 a+100 b+10 c+b$.
According to the problem, we have
$$
\begin{aligned}
& (1000 c+100 b+10 a+b)-(1000 a+100 b+10 c+b) \\
= & 5940 .
\end{aligned}
$$

That is
$$
\begin{array}{l}
990(c-a)=5940, \\
c-a=6 .
\end{array}
$$

Since we are looking for the smallest odd number, $a$ should be as small as possible. We take $a=1$, then $c=$
7.

At this point, $a+c=8$, and since the four-digit number leaves a remainder of 8 when divided by 9, $a+b+c+b$ leaves a remainder of 8 when divided by 9, so $2 b$ is divisible by 9.
Thus,
$$
b=9 \text { or } b=0 .
$$

Since the number we are looking for is odd, $b=9$. Therefore, the four-digit number we are looking for is 1979.