II. (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 $x$.

保留源文本的换行和格式，直接输出翻译结果。
Second, first prove: "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. Then, for any integer $n, n a$ is an integer. Therefore, $n a=[n a]$. Thus,
$$
\begin{array}{l}
{[n c]=[n(a+b)]=[n a+n b]} \\
=n a+[n b]=[n a]+[n b] .
\end{array}
$$

On the other hand, if for any integer $n$, $[n a]+[n b]=[n c]$. Then by taking $n=1, -1$, we get
$$
[a]+[b]=[c],[-a]+[-b]=[-c] \text{. }
$$

Adding the two equations, we get
$$
[a]+[-a]+[b]+[-b]=[c]+[-c] \text{. }
$$

For any real number $x$,
$$
\begin{array}{l}
{[x]+[-x]=[x]+[-([x]+\{x\})]} \\
=[x]+[-[x]-\{x\}] \\
=[-\{x\}]=\left\{\begin{array}{cc}
0, & x \in \mathbf{Z} ; \\
-1, & x \notin \mathbf{Z} .
\end{array}\right.
\end{array}
$$

Thus, if $a, b$ are not integers, then
$$
\begin{array}{l}
{[a]+[-a]=[b]+[-b]=-1 .} \\
\text{ Hence }[a]+[-a]+[b]+[-b] \\
=-2 \neq[c]+[-c],
\end{array}
$$

which is a contradiction.
Therefore, $a, b$ at least one is an integer. Without loss of generality, assume $a$ is an integer. Then, for any integer $n, n a$ is an integer. Thus, $n a=[n a]$. Then for any integer $n$,
$$
\begin{array}{l}
{[n c]=[n a]+[n b]=n a+[n b]} \\
=[n a+n b]=[n(a+b)],
\end{array}
$$

i.e.,
$$
\begin{array}{l}
n c-\{n c\}=n(a+b)-\{n(a+b)\} . \\
\text{ Hence } a+b-c=\frac{\{n(a+b)\}-\{n c\}}{n} . \\
\text{ And } \{n(a+b)\}-\{n c\} \mid<2, \text{ thus, } \\
a+b-c=\lim _{n \rightarrow \infty}(a+b-c) \\
=\lim _{n \rightarrow \infty} \frac{\{n(a+b)\}-\{n c\}}{n}=0 .
\end{array}
$$

In summary, the required real number triplet $(a, b, c)=$ $(m, t, m+t),(t, n, n+t),(m, n, m+n)$, where $m, n$ are any integers, and $t$ is any real number.