Three. (20 points) Let $a, b$ be integers. How many solutions does the system of equations
$$
\left\{\begin{array}{l}
{[x]+2 y=a,} \\
{[y]+2 x=b}
\end{array}\right.
$$
(here $[x]$ denotes the greatest integer less than or equal to $x$) have?
Obviously, if $2x$ and $2y$ are integers, then $x$ and $y$ are either integers or differ from some integer by $\frac{1}{2}$. We will discuss several cases below:

(1) Suppose $x, y$ are both integers, let $x=[x], y=[y]$, then from the system of equations we can obtain
$$
x=\frac{2b-a}{3}, \quad y=\frac{2a-b}{3}.
$$

The system has a solution if and only if $3 \mid (2b-a)$ and $3 \mid (2a-b)$, i.e., $3 \mid (a+b)$.

(2) Suppose $x$ is not an integer, but $y$ is an integer, then from the system of equations we can solve for
$$
x=\frac{2b-a-2}{3}+\frac{1}{2}, \quad y=\frac{2a-b+1}{3}.
$$

The system has a solution if and only if $3 \mid (2b-a-2)$ and $3 \mid (2a-b+1)$, i.e., $3 \mid (a+b-1)$.

(3) Suppose $x$ is an integer, but $y$ is not an integer, then similarly we can solve for
$$
x=\frac{2b-a+1}{3}, \quad y=\frac{2a-b-2}{3}+\frac{1}{2}.
$$

The system has a solution if and only if $3 \mid (2b-a+1)$ and $3 \mid (2a-b-2)$, i.e., $3 \mid (a+b-1)$.

(4) Suppose both $x$ and $y$ are not integers, then
$$
x=\frac{2b-a-1}{3}+\frac{1}{2}, \quad y=\frac{2a-b-1}{3}+\frac{1}{2}.
$$

The system has a solution if and only if $3 \mid (2b-a-1)$ and $3 \mid (2a-b-1)$, i.e., $3 \mid (a+b-2)$.

In summary, when $3 \mid (a+b-1)$, there are two solutions; when $3 \mid (a+b)$ and $3 \mid (a+b-2)$, there is a unique solution.