$20\{x\} = x + [x] + [x + 0.5]$

Solution 1:

Let call $n=\lfloor x\rfloor\in\mathbb Z$ and $r=\{x\}\in[0,1)$ to avoid writing all these brackets.

$x\in A\iff 20r=n+r+n+\lfloor n+r+0.5\rfloor=3n+r+\lfloor r+0.5\rfloor$

• for $r<0.5$ we have $19r=3n\implies n=\frac {19r}3<\frac{19}6\implies n\le 3$
• for $r\ge0.5$ we have $19r=3n+1\implies \frac{19-2}6\le n=\frac{19r-1}3<\frac{19-1}3\implies 3\le n\le 5$

Since there are so few $n$ possible, let calculate the corresponding $x$:

• For small $r$ first $n=0,1,2,3$ then $r=0,\frac 3{19},\frac 6{19},\frac 9{19}$
• For big $r$ now $n=3,4,5$ then $r=\frac {10}{19},\frac {13}{19},\frac {16}{19}$

$S=(0+1+2+3+3+4+5)+\dfrac 1{19}(0+3+6+9+10+13+16)=18+\dfrac{57}{19}=21$

I let you find $[S]$.

Solution 2:

Rewriting the constraint $$19 \{x\} = [x] + [x] + [x + 0.5]$$ we see, that left part is in range $[0; 19]$, and right part is integer. Let that integer be $k$.

Now we can through all $k=0, 1, ..., 19$, and find $x$ for each of them (of course, not for all $k$ such $x$ exists). It's straightforward to express $\{x\}$ and $[x]$ in terms of $k$:

$$ \{x\} = \frac{k}{19} \\ k = [x] + [x] + [x + 0.5] = 2 [x] + [ [x] + \{x\} + 0.5] = \\ = 3 [x] + [\{x\} + 0.5] = 3 [x] + \left[ \frac{k}{19} + \frac12 \right] \\ [x] = \frac13 \left( k - \left[ \frac{k}{19} + \frac12 \right]\right) $$

Now with some diligence (which I lack) we can find all $x$ in $A$ and get the answer.