Four, (50 points) Find all real triples $(a, b, c)$ such that for any integer $n$, we have $[n a]+[n b]=[n c]$. Here $[x]$ denotes the greatest integer not exceeding the real number $x$.

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Four, Solution: First, we prove that "for any integer $n$, $[n a]+[n b]=[n c]$" is equivalent to "$a, b$ at least one is an integer and $c=a+b$."

On one hand, if $a, b$ at least one is an integer and $c=a+b$, then without loss of generality, assume $a$ is an integer. For any integer $n$, $n a$ is an integer, so $[n a]=n a$. Therefore, $[n c]=[n(a+b)]=[n a]+[n b]$.

On the other hand, if for any integer $n$, $[n a]+[n b]=[n c]$. Then take $n=1, -1$,
we get $[a]+[b]=[c]$, $[-a]+[-b]=[-c]$. Adding these two equations, we get $[a]+[-a]+[b]+[-b]=[c]+[-c]$.
For any real number $x$, $[x]+[-x]=[x]+[-([x]+\{x\})]=[-\{x\}]=\left\{\begin{array}{l}0, x \in \mathbf{Z} \\ -1, x \notin \mathbf{Z}\end{array}\right.$.
If $a, b$ are not integers, then $[a]+[-a]+[b]+[-b]=-2 \neq [c]+[-c]$, which is a contradiction.
Therefore, $a, b$ at least one is an integer. Without loss of generality, assume $a$ is an integer, then $[n c]=[n a]+[n b]=n a+[n b]=[n a+n b]$ $=[n(a+b)]$, i.e.,
$$
n c-\{n c\}=n(a+b)-\{n(a+b)\} \text{. Hence } a+b-c=\frac{\{n(a+b)\}-\{n c\}}{n} \text{. }
$$

Since $|\{n(a+b)\}-\{n c\}|<2$, we have $a+b-c=\lim _{n \rightarrow \infty}(a+b-c)=\lim _{n \rightarrow \infty} \frac{\{n(a+b)\}-\{n c\}}{n}=0$.
In summary, the required real number triples $(a, b, c)=(m, t, m+t),(t, n, n+t),(m, n, m+n)$, where $m, n$ are any integers, and $t$ is any real number.