Three. (50 points)
Find all non-military integers $a, b, a \neq b$, such that:
the set of integers can be partitioned into 3 subsets, so that for each $n, n+a, n+b$ belong to these 3 subsets respectively.
Three, (50 points)
Let $(a, b)=r$. Consider $a^{\prime}=\frac{a}{r}, b^{\prime}=\frac{b}{r}$, then $a^{\prime} \cdot b^{\prime}=2(\bmod 3)(*)$ is the required necessary and sufficient condition.
First, prove sufficiency. When $a^{\prime} \cdot b^{\prime} \equiv 2(\bmod 3)$, consider the following 3 sets:
$A_{i}\{n \mid n=k r+t, 0 \leqslant t \leqslant r-1$, and $k \equiv i(\bmod 3)\}(i=1,2,3)$, then the set of natural numbers is partitioned into 3 sets $A_{1}, A_{2}, A_{3}$.
Next, prove that this partition satisfies the condition.
Let $n=k r+t \quad(0 \leqslant t \leqslant r-1)$,
then $n+a=\left(k+a^{\prime}\right) r+t, n+b=\left(k+b^{\prime}\right) r+t$.
Since $a^{\prime} b^{\prime} \equiv 2(\bmod 3)$. Therefore, $0, a^{\prime}, b^{\prime}$ form a complete residue system modulo 3, so $k, k+a^{\prime}, k+b^{\prime}$ form a complete residue system modulo 3, so $n, n+a, n+b$ are in different subsets.

Next, prove necessity. Consider $n=m^{\prime}\left(n^{\prime} \in \mathbb{Z}\right)$, then $m^{\prime}, m^{\prime}+r^{\prime}, m^{\prime}+b^{\prime}$ belong to different subsets.
Let $m^{\prime} \in A_{1}, m^{\prime}+r a^{\prime} \in A_{2}, m^{\prime}+r b^{\prime} \in A_{3}$,
and $r n^{\prime}+r a^{\prime}+r b^{\prime} \in A_{2}, A_{3}$, so $r\left(n^{\prime}+a^{\prime}+b^{\prime}\right) \in A_{1}$,
also $r\left(n^{\prime}+2 a^{\prime}\right) \in A_{1}, A_{2}$, so $r\left(n^{\prime}+2 a^{\prime}\right) \in A_{3}$.
Similarly, we get: $r\left(n^{\prime}+3 a^{\prime}\right) \in A_{1}$, thus $r\left(n^{\prime}+3 l a^{\prime}\right) \in A_{1},(l \in \mathbb{Z})$.
Similarly, $r\left(n^{\prime}+3 k b^{\prime}\right) \in A_{1}$, thus $r\left(n^{\prime}+3 l a^{\prime}+3 k b^{\prime}\right) \in A_{1},(l, k \in \mathbb{Z})$.
If $3 \mid a^{\prime}$, then by Bézout's theorem, there exist $k, l \in \mathbb{Z}$, such that $l a^{\prime}+k b^{\prime}=\frac{a^{\prime}}{3}$, thus $r\left(n^{\prime}+a^{\prime}\right) \in A_{1} \cap A_{2}=\emptyset$, a contradiction! Therefore, $3 \nmid a^{\prime}$. Similarly, $3 \nmid b^{\prime}, 3 \nmid a^{\prime}-b^{\prime}$. Thus, $0, a^{\prime}, b^{\prime}$ are distinct modulo 3, so $a^{\prime} b^{\prime} \equiv 2(\bmod 3)$. The necessity is proved.