Example 4 Given that $a$ is a positive integer, and makes the quadratic equation in $x$
$$
a x^{2}+2(2 a-1) x+4(a-3)=0
$$

have at least one integer root. Find the value of $a$.


Solution: Transform the original equation into
$$
(x+2)^{2} a=2(x+6) \text {. }
$$

Obviously, $x+2 \neq 0$, thus,
$$
a=\frac{2(x+6)}{(x+2)^{2}} .
$$

Since $a$ is a positive integer, then $a \geqslant 1$, i.e.,
$$
\begin{array}{l}
\frac{2(x+6)}{(x+2)^{2}} \geqslant 1 \Rightarrow x^{2}+2 x-8 \leqslant 0 \\
\Rightarrow(x+4)(x-2) \leqslant 0,
\end{array}
$$

Therefore, $-4 \leqslant x \leqslant 2(x \neq-2)$.
When $x=-4,-3,-1,0,1,2$, we get $a$ as $1,6,10,3, \frac{14}{9}, 1$.
Since $a$ is an integer, the values of $a$ are $1,3,6,10$.
Note: From the solution process, when $a=1$, the equation has two integer roots $-4,2$; when $a=3,6,10$, the equation has only one integer root. Sometimes, in a quadratic equation in $x$, if the parameter is linear, we can first solve for this parameter. This problem can also be solved using the discriminant. That is,
$$
\Delta=4\left[(2 a-1)^{2}-4 a(a-3)\right]=4(8 a+1)
$$

is a perfect square, so $8 a+1$ is a square number, and it is the square of an odd number.
Let $8 a+1=(2 m+1)^{2}, m$ is a positive integer, then
$$
a=\frac{m^{2}+m}{2} \text {. }
$$

Thus, the original equation can be transformed into
$$
\begin{array}{l}
m(m+1) x^{2}+4\left(m^{2}+m-1\right) x+ \\
4(m-2)(m+3)=0,
\end{array}
$$

i.e., $[m x+2(m-2)][(m+1) x+2(m+3)]=0$.
Solving, we get $x_{1}=-2+\frac{4}{m}, x_{2}=-2-\frac{4}{m+1}$.
Therefore, $m \mid 4$ or $(m+1) \mid 4$.
Hence, $m=1,2,4$ or $m=1,3$.
Thus, the values of $a$ are $1,3,6,10$.