Lian Example 18 (2005 National Training Team Test Question) Find all positive integers $m, n$, such that the inequality
$\left[(m+n)_{\alpha}\right]+[(m+n) \beta] \geqslant[m \alpha]+[m \beta]+[n(\alpha+\beta)]$
holds for any real numbers $\alpha, \beta$. Here $[x]$ denotes the integer part of the real number $x$.
The answer is $m=n$.
If $m=n$, then the original inequality becomes
$$
[2 m \alpha]+[2 m \beta] \geqslant[m \alpha]+[m \beta]+[m(\alpha+\beta)],
$$

Let $x=m \alpha, y=m \beta$, then the above inequality becomes
$$
[2 x]+[2 y] \geqslant[x]+[y]+[x+y],
$$

which is equivalent to $[2\{x\}]+[2\{y\}] \geqslant[\{x\}+\{y\}]$.
Assume without loss of generality that $x \geqslant y$, then $[2\{x\}]+[2\{y\}] \geqslant[2\{x\}] \geqslant[\{x\}+\{y\}]$, thus (2) holds.
Conversely, suppose the inequality (1) holds for all $\alpha, \beta$, take $\alpha=\beta=\frac{1}{m+n+1}$, we get $\left[\frac{2 n}{m+n+1}\right] \leqslant 0$, hence $2 n<m+n+1$, i.e., $n \leqslant m$.

On the other hand, let $d=(m, n)$, by Bézout's identity, we can take non-negative integers $x, y$ such that $n x=m y+$ $m-d$ [we can first take positive integers $x^{\prime}, y^{\prime}$, such that $m y^{\prime}-n x^{\prime}=d \Rightarrow m\left(y^{\prime}-1\right)-n x^{\prime}=d-m$ ]. In (1), take $\alpha=\beta=\frac{x}{m}$, then
$$
2\left[\frac{x n}{m}\right] \geqslant\left[\frac{2 x n}{m}\right],
$$

thus $2\left\{\frac{x n}{m}\right\} \leqslant\left\{\frac{2 x n}{m}\right]<1$,
so $\left\{\frac{x n}{m}\right\}<\frac{1}{2}$,
i.e., $\left\{\frac{m y+m-d}{m}\right\}<\frac{1}{2}$,
i.e., $\frac{m-d}{m}<\frac{1}{2}$,
thus $m<2 d$.
Since $d \mid m$, hence $m=d$ (if $m=d q$, then by $d q<2 d$ we know $q=1$ ), so $m \mid n$, thus $m \leqslant n$.
In summary, we have $m=n$.