Three. (25 points) Given that $a$ and $b$ are any real numbers, find the minimum value of the maximum number among $|2a+b|$, $|2a-b|$, and $|1-b|$.
Three, let the largest number among $|2a + b|$, $|2a - b|$, and $|1 - b|$ be $x$. Then,
\[ |2a + b| \leq x, \quad |2a - b| \leq x, \quad |1 - b| \leq x. \]
Thus,
\[ -x \leq 2a + b \leq x, \]
\[ -x \leq b - 2a \leq x, \]
\[ -x \leq 1 - b \leq x. \]
Adding (1) + (2) + 2 \times (3) gives:
\[ -4x \leq 2 \leq 4x. \]
Solving this, we get:
\[ x \geq \frac{1}{2}. \]
When $2a + b = \frac{1}{2}$, $b - 2a = \frac{1}{2}$, and $1 - b = \frac{1}{2}$, we get $a = 0$, $b = \frac{1}{2}$. At this point,
\[ |2a + b| = |2a - b| = |1 - b| = \frac{1}{2}, \quad x = \frac{1}{2}. \]

Therefore, the minimum value of the largest number among $|2a + b|$, $|2a - b|$, and $|1 - b|$ is $\frac{1}{2}$.