Question: How should I find the number of subsets of $[n]$ like $A$ where $\max A - \min A$ is an odd number?  

I tried to solve the problem for a simpler example but it didn't help.

Answer: 
There are $\max A-\min A-1$ numbers in between that may or may not be present, so $2^{\delta-1}$ different subsets for a given value of $\delta=\max A-\min A$. For given $n$, there are $n-\delta$ pairs in $[n]$ with difference $\delta$, so the desired number is

$$
\sum\_{k=0}^{l-1}(n-(2k+1))2^{(2k+1)-1}=\sum\_{k=0}^{l-1}(n-(2k+1))4^k=\frac19\left(4^l(3n-6l+5)-3n-5\right)
$$

with the limit $l=\lfloor n/2\rfloor$.

This simplifies to $\frac19\left(r\cdot2^n-3n-5\right)$, with $r=5$ for even $n$ and $r=4$ for odd $n$.
How do we create such an $A$? Choose the largest element of $A$ call it $M$. Then we choose the smallest element, $m$, of $A$ such that $M - m$ is odd. Now to round out $A$, we choose among the remaining $M-m-1$ elements between $M$ and $m$. There are $2^{M-m-1}$ ways to do this. Thus the total number of subsets, $A$, of $[n]$ with the desired property is 
\begin{align\*} \sum\_{M=2}^n \sum\_{\substack{m < M \\ M-m \textrm{ odd}}} 2^{M-m-1} &= \sum\_{M=2}^n \sum\_{k=0}^{\lfloor\frac{M}{2}\rfloor} 2^{2k+1-1}\\
&=\sum\_{M=2}^n \sum\_{k=0}^{\lfloor\frac{M}{2}\rfloor} 4^k\\
&=\sum\_{M=2}^n \frac{4^{\lfloor\frac{M}{2}\rfloor}-1}{4-1}\\
&=\frac{1}{3} \sum\_{M=2}^n (4^{\lfloor\frac{M}{2}\rfloor}-1)
\end{align\*}

I'll leave the rest to you. I'm sure this could be simplified further.