3. Given $A\left(x_{1}, y_{1}\right)$ and $B\left(x_{2}, y_{2}\right)$ are
points on the circle $\odot O: x^{2}+y^{2}=1$,
and $|A B|=1$. Then the maximum value of $\left|x_{1}+y_{1}-1\right|+$ $\left|x_{2}+y_{2}-1\right|$ is $\qquad$
3. $2+\sqrt{6}$.

From the problem, we know that $\triangle O A B$ is an equilateral triangle.
Then, the geometric meaning of $\frac{\left|x_{1}+y_{1}-1\right|}{\sqrt{2}}+\frac{\left|x_{2}+y_{2}-1\right|}{\sqrt{2}}$ 
is the sum of the distances from points $A$ and $B$ to the line $x+y-1=0$, and its maximum value is twice the distance from the midpoint $M$ of $A B$ to the line $x+y=1$. The trajectory of the midpoint $M$ is a circle with the origin $O$ as the center and $\frac{\sqrt{3}}{2}$ as the radius, so the maximum distance from point $M$ to the line $x+y=1$ is $\frac{\sqrt{2}+\sqrt{3}}{2}$, that is,
$$
\left(\frac{\left|x_{1}+y_{1}-1\right|}{\sqrt{2}}+\frac{\left|x_{2}+y_{2}-1\right|}{\sqrt{2}}\right)_{\max }=\sqrt{2}+\sqrt{3} \text {. }
$$

Therefore, the required value is $2+\sqrt{6}$.