Solution: It follows from $|4 a-2| \leq 1$ that $\frac{1}{4} \leq a \leq \frac{3}{4}$. Also, it is clear that when $a \geq \frac{1}{4}$, then $A \geq 0$ and $A=0$ only if $a=\frac{1}{4}$. Let $k$ be a positive integer such that $A=k$. Then $27 a^{4}-4 a l+l=0$, where $l=\frac{1}{k}$. Multiply the above equality by 3 and write it in the following way:

$$\begin{aligned} & 81 a^{4}-18 a^{2}+1+18 a^{2}-12 a l+3 l-1=0 \Longleftrightarrow \\ & \left(9 a^{2}-1\right)^{2}+2(3 a-1)^{2}-12 a(l-1)+3(l-1)=0 \end{aligned}$$

so $\left(9 a^{2}-1\right)^{2}+2(3 a-1)^{2}+3(l-1)(1-4 a)=0$. Therefore $3(l-$ $1)(1-4 a) \leq 0$, which is possible (recall $a>\frac{1}{4}$ ) only if $l \geq 1$ or $k \leq 1$. But since $k$ is a positive integer, it follows that $k=1$ and $a=\frac{1}{3}$. Therefore the required values are $a=\frac{1}{4}$ and $a=\frac{1}{3}$.