T10. Find all positive integers $a$ such that the quadratic equation $a x^{2}+2(2 a-1) x+4(a-3)=0$ has at least one integer root. (3rd Zu Chongzhi Cup Junior High School Mathematics Invitational Competition)
Solving, the original equation becomes $a=\frac{2 x+12}{(x+2)^{2}}$. Since $a$ is a positive integer, we have $a \geq 1$, i.e., $\frac{2 x+12}{(x+2)^{2}} \geq 1$. Simplifying, we get $x^{2}+2 x-8 \leqslant 0$. Solving this, we get $-4 \leqslant x \leqslant 2$.

Since $x$ is an integer root and clearly $x \neq-2$, therefore, $x$ can only take the values $-4,-3,-1,0,1,2$.
Substituting into the original equation (or the expression for $a$) we get
$$
\begin{array}{l}
\left\{\begin{array} { l } 
{ x = - 4 , } \\
{ a = 1 , }
\end{array} \left\{\begin{array}{l}
x=-3, \\
a=6,
\end{array},\left\{\begin{array}{l}
x=-1, \\
a=10,
\end{array}\right.\right.\right. \\
\left\{\begin{array} { l } 
{ x = 0 , } \\
{ a = 3 , }
\end{array} \left\{\begin{array}{l}
x=1, \\
a=\frac{14}{9},
\end{array},\left\{\begin{array}{l}
x=2, \\
a=1,
\end{array}\right.\right.\right.
\end{array}
$$

In summary, there are 4 positive integer values of $a$ that satisfy the requirements, namely $a=1,3,6,10$.