The graph of the equation \[\sqrt{x^2+y^2} + |y-1| = 3\]consists of portions of two different parabolas. Compute the distance between the vertices of the parabolas.
We take cases on the sign of $y-1.$ If $y \ge 1,$ then the equation simplifies to \[\sqrt{x^2 + y^2} + (y-1) = 3,\]or \[\sqrt{x^2+y^2} = 4-y.\]Squaring both sides, we get $x^2 + y^2 = (4-y)^2 = y^2 -8y + 16,$ or $x^2 = -8y + 16.$ Solving for $y,$ we get \[y = -\frac{1}{8}x^2 + 2,\]so the vertex of this parabola is $(0, 2).$

If $y < 1,$ then we have \[\sqrt{x^2+y^2} + (1-y) = 3,\]or \[\sqrt{x^2+y^2} = y+2.\]Squaring both sides, we get $x^2+y^2 = y^2+4y+4,$ and solving for $y$ gives \[y = \frac14x^2-1.\]So the vertex of this parabola is $(0, -1).$

Thus, the distance between the vertices of the two parabolas is $|2 - (-1)| = \boxed{3}.$ (Note that $3$ is the number that appears on the right-hand side. Is this a coincidence?)