There is a solution, I can only give an answer: $0$, $\frac{1}{3}$, $\frac{2}{3}$.

If we calculate $\mod 1$, we get $\{x\} \equiv x$. But then $3\{x+y+z+t\}\equiv 3x+3y+3z+3t \equiv (x+y+z)+(x+y+t)+(x+z+t)+(y+z+t) \equiv \{x+y+z\}+\{x+y+t\}+\{x+z+t\}+\{y+z+t\} \equiv 0 \mod 1$ and thus, because we also have ${x+y+z+t} \in [0;1)$, we know that $0, \frac{1}{3}$ and $\frac{2}{3}$ are the only possible solutions. Now, we can set all variables on $\frac{1}{12}, \frac{5}{8}$ and $\frac{3}{4}$ to see that these 3 values are indeed possible.

$ x+y+z -[x+y+z]=y+z+t-[y+z+t]\\ \ x- [x+y+z]= t - [y+z+t] \\ $ ( [x] - {x}) - [x+y+z] =([t] - {t} ) - [y+z+t] thus {x}={t} which gives us that {x}={y}={z}={t} Next {3x}=1/4 3x-[3x]=1/4 3([x]-{x})-[3x]=1/4 Thus 3{x}=1-1/4 or 2-1/4 or 3-1/4 so {x}=1/4, 7/12, 11/12 and {4x}=0, 1/3, 2/3.

sedrikktl wrote: $ ( [x] - {x}) - [x+y+z] =([t] - {t} ) - [y+z+t] $ I think here should be ( [x]+ {x}) - [x+y+z] =([t] + {t}) - [y+z+t]

yes, my bad. But it does not matter so far.