Suppose that for real $x,y,z,t$ the following equalities hold:$\{x+y+z\}=\{y+z+t\}=\{z+t+x\}=\{t+x+y\}=1/4$.
Find all possible values of $\{x+y+z+t\}$.(Here$\{x\}=x-[x]$)
1. Given the equalities:
   \[
   \{x+y+z\} = \{y+z+t\} = \{z+t+x\} = \{t+x+y\} = \frac{1}{4}
   \]
   where \(\{x\} = x - \lfloor x \rfloor\) denotes the fractional part of \(x\).

2. We need to find the possible values of \(\{x+y+z+t\}\).

3. Consider the given equalities modulo 1:
   \[
   \{x+y+z\} \equiv x+y+z \pmod{1}
   \]
   \[
   \{y+z+t\} \equiv y+z+t \pmod{1}
   \]
   \[
   \{z+t+x\} \equiv z+t+x \pmod{1}
   \]
   \[
   \{t+x+y\} \equiv t+x+y \pmod{1}
   \]

4. Since \(\{x+y+z\} = \{y+z+t\} = \{z+t+x\} = \{t+x+y\} = \frac{1}{4}\), we have:
   \[
   x+y+z \equiv \frac{1}{4} \pmod{1}
   \]
   \[
   y+z+t \equiv \frac{1}{4} \pmod{1}
   \]
   \[
   z+t+x \equiv \frac{1}{4} \pmod{1}
   \]
   \[
   t+x+y \equiv \frac{1}{4} \pmod{1}
   \]

5. Adding these four congruences:
   \[
   (x+y+z) + (y+z+t) + (z+t+x) + (t+x+y) \equiv 4 \cdot \frac{1}{4} \pmod{1}
   \]
   \[
   3(x+y+z+t) \equiv 1 \pmod{1}
   \]

6. Simplifying the above congruence:
   \[
   3(x+y+z+t) \equiv 0 \pmod{1}
   \]

7. Since \(x+y+z+t\) is a real number, we can write:
   \[
   3\{x+y+z+t\} \equiv 0 \pmod{1}
   \]

8. The possible values of \(\{x+y+z+t\}\) that satisfy this congruence are:
   \[
   \{x+y+z+t\} = 0, \frac{1}{3}, \frac{2}{3}
   \]

9. To verify, we can set specific values for \(x, y, z, t\) such that the given conditions hold. For example:
   - If \(x = \frac{1}{12}, y = \frac{1}{12}, z = \frac{1}{12}, t = \frac{1}{12}\), then:
     \[
     x+y+z+t = \frac{1}{12} + \frac{1}{12} + \frac{1}{12} + \frac{1}{12} = \frac{1}{3}
     \]
     \[
     \{x+y+z+t\} = \frac{1}{3}
     \]

   - If \(x = \frac{1}{4}, y = \frac{1}{4}, z = \frac{1}{4}, t = \frac{1}{4}\), then:
     \[
     x+y+z+t = \frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = 1
     \]
     \[
     \{x+y+z+t\} = 0
     \]

   - If \(x = \frac{1}{6}, y = \frac{1}{6}, z = \frac{1}{6}, t = \frac{1}{6}\), then:
     \[
     x+y+z+t = \frac{1}{6} + \frac{1}{6} + \frac{1}{6} + \frac{1}{6} = \frac{2}{3}
     \]
     \[
     \{x+y+z+t\} = \frac{2}{3}
     \]

The final answer is \( \boxed{ \{x+y+z+t\} = 0, \frac{1}{3}, \frac{2}{3} } \).