10. Let $x$ be a real number, and define $\lceil x\rceil$ as the smallest integer not less than the real number $x$ (for example, $\lceil 3.2 \rceil = 4, \lceil -\pi \rceil = -3$). Then, the sum of all real roots of the equation
$$
\lceil 3 x+1\rceil=2 x-\frac{1}{2}
$$

is equal to
10. -4 .

Let $2 x-\frac{1}{2}=k \in \mathbf{Z}$. Then
$$
x=\frac{2 k+1}{4}, 3 x+1=k+1+\frac{2 k+3}{4} \text {. }
$$

Thus, the original equation is equivalent to
$$
\begin{array}{l}
{\left[\frac{2 k+3}{4}\right]=-1 \Rightarrow-2<\frac{2 k+3}{4} \leqslant-1} \\
\Rightarrow-\frac{11}{2}<k \leqslant-\frac{7}{2} \\
\Rightarrow k=-5 \text { or }-4 .
\end{array}
$$

The corresponding $x$ values are $-\frac{9}{4},-\frac{7}{4}$.
Therefore, the sum of all real roots is -4.