$12 \cdot 4$ Let $a, b$ be integers, what is the smallest positive integer value $c$ of $\Delta=\left|\begin{array}{ll}36 & a \\ 81 & b\end{array}\right|$? For each group of positive integer solutions $(a, b)$ that satisfies $\Delta=c$, to make $a+b$ the smallest, what are $a, b$?
(Shanghai Mathematical Competition, China, 1963)
[Solution]
$$
\begin{aligned}
\Delta & =36 b-81 a \\
& =9(4 b-9 a)
\end{aligned}
$$

Since $a, b$ are integers, $4 b-9 a$ is also an integer.
Since $(4,9)=1$, there exist integers $a, b$ such that
$$
4 b-9 a=1 .
$$

Thus, the smallest positive integer value of $\Delta$ is $c=9$.
Solving equation (1) gives
$$
\left\{\begin{array}{l}
a=4 t-1 \\
b=9 t-2 .
\end{array}\right.
$$

When $t=1,2,3, \cdots$, the above equations provide all positive integer solutions.
$$
a+b=13 t-3 \quad(t=1,2,3, \cdots) .
$$

Therefore, when $t=1$, the value of $a+b$ is the smallest, with the minimum value being 10. At this time, $a=3, b=7$.