10 Let $x$ be a real number, and define $\lceil x\rceil$ as the smallest integer not less than $x$, for example $\lceil\pi\rceil=4$, $\lceil-\pi\rceil=-3$. The sum of all real roots of the equation $\lceil 3 x+1\rceil=2 x-\frac{1}{2}$ is $\qquad$ .
(10) -4 Hint: Let $2 x-\frac{1}{2}=k \in \mathbf{Z}$, then $x=\frac{2 k+1}{4}, 3 x+1=k+1+\frac{2 k+3}{4}$, so the original equation is equivalent to $\left\lceil\frac{2 k+3}{4}\right\rceil=-1$, that is,
$$
-2<\frac{2 k+3}{4} \leqslant-1,
$$

thus $-\frac{11}{2}<k \leqslant-\frac{7}{2}$, i.e., $k=-5$ or -4.
The corresponding $x$ values are $-\frac{9}{4},-\frac{7}{4}$. Therefore, the sum of all real roots is -4.