1. Let $[x]$ be the greatest integer not exceeding the real number $x$. For any $x \in \mathbf{R}$, we have $[x]+[x+a]=[2 x]$. Then the set of all possible values of the positive real number $a$ is
-1. $\left\{\frac{1}{2}\right\}$.
Let $x=\frac{1}{2}$. Then
$$
\left[\frac{1}{2}+a\right]=1 \Rightarrow a \in\left[\frac{1}{2}, \frac{3}{2}\right) \text {. }
$$

For any $x \in\left(0, \frac{1}{2}\right)$, we always have
$$
[x+a]=0 \Rightarrow x+a<1,
$$

which means $a<1-x$ always holds.
Therefore, $a \leqslant \frac{1}{2}$. Hence $a=\frac{1}{2}$.
When $a=\frac{1}{2}$, for any $x \in \mathbf{R}$, we have
$$
[x]+[x+a]=[2 x] .
$$