14.78 Determine all pairs of real numbers $(a, b)$ such that $a[b n]=b[a n]$ for all positive integers $n$.
(39th International Mathematical Olympiad Preliminary Question, 1998)
[Solution] Let $n=1$, we get $a[b]=b[a]$,
Let $a=[a]+\{a\}, b=[b]+\{b\}$, where $\{a\},\{b\}$ are the fractional parts of $a, b$, then from the given we have
$$
\begin{array}{c}
a[[b] n+\{b\} n]=b[[a] n+\{a\} n], \\
a[b] n+a[\{b\} n]=b[a] n+b[\{a\} n],
\end{array}
$$

From (1) we get $\quad a[\{b\} n]=b[\{a\} n]$.
There exists a natural number $k$, such that $\{a\}<\frac{1}{k}$, let $n=k$, from (2) we get
$$
a[k\{b\}]=0 .
$$

If $a \neq 0$, then from (3) we can get $\{b\}<\frac{1}{k}$. This indicates that when $a$ is an integer (i.e., $\{a\}=0$), $b$ must also be an integer.
Similarly, if $b \neq 0$, and $\{b\}<\frac{1}{k}$, then $\{a\}<\frac{1}{k}$.
Thus, when $a, b$ are not integers, there exists $k$, such that
$$
\frac{1}{k+1} \leqslant\{a\}<\frac{1}{k}, \quad \frac{1}{k+1} \leqslant\{b\}<\frac{1}{k} \text {. }
$$

Let $n=k+1$, then
$$
1 \leqslant n\{b\}<\frac{k+1}{k}<2 \quad 1 \leqslant n\{a\}<\frac{k+1}{k}<2
$$

From (2) we get $a=b$.
Therefore, the solutions to this problem are
$(0, a),(a, 0),(a, a)$ (when $a$ is any real number)
$(a, b)$ (when $a, b$ are both integers).