Let a=[a]+a=n+f, where $n=[a]\in \mathbb Z$ and $f=\{a\}\in[0,1)$.

Then $2a-1=2(n+f)-1=2n+2f-1,$ and the given equation becomes $|2n+2f-1|=3n+2f.$

We solve this by cases.

1. Necessary condition

Since the left side is non-negative, $3n+2f\ge 0.$ Because $0\le f<1$, this restricts possible integers $n$.

2. Case I: $2n+2f-1\ge 0$

Then $|2n+2f-1|=2n+2f-1.$ So $2n+2f-1=3n+2f$ $\Rightarrow -1=n.$ Thus $n=-1.$ But for this case we also need $2(-1)+2f-1\ge 0 \Rightarrow 2f-3\ge 0 \Rightarrow f\ge \frac32,$ which is impossible since $f<1$.

So no solution from this case.

3. Case II: $2n+2f-1<0$

Then $|2n+2f-1|=-(2n+2f-1)=-2n-2f+1.$ Hence $-2n-2f+1=3n+2f$ $\Rightarrow 1=5n+4f$ $\Rightarrow f=\frac{1-5n}{4}.$

Now use $0\le f<1$: $0\le \frac{1-5n}{4}<1.$ Multiply by $4$: $0\le 1-5n<4.$ This gives $1-5n\ge 0 \Rightarrow n\le \frac15,$ and $1-5n<4 \Rightarrow -5n<3 \Rightarrow n> -\frac35.$ Since $n\in\mathbb Z$, the only possibility is $n=0.$ Then $f=\frac{1-0}{4}=\frac14.$ So $a=n+f=\frac14.$ Check: $|2a-1|=\left|\frac12-1\right|=\frac12,$ $3[a]+2\{a\}=3(0)+2\left(\frac14\right)=\frac12.$ Correct.

Thus $S=\left\{\frac14\right\}.$

4. Required value

$\sum_{a\in S} a=\frac14.$ Therefore, $72\sum_{a\in S} a=72\cdot \frac14=18.$

5. Comparison with stored answer

Our derived answer is $18$, which matches the stored correct answer.