Firstly, suppose that $a$ and $b$ are not integers and let $a_1$, $b_1$ be the fractional part of $a$ and $b$ respectively. Note that for $n=1$, we have: $$\frac{a}{b}=\frac{\lfloor a \rfloor}{\lfloor b \rfloor}=\frac{\lfloor a \cdot n \rfloor}{\lfloor b \cdot n \rfloor}=\frac{p}{q}$$where $\frac{p}{q}$ is the rational number $\frac{a}{b}$ written in lowest terms. WLOG $p>0$ and assume that $p>1$. Then $p$ must divide $\lfloor a \cdot n \rfloor$ for all $n$. Let $n$ be the smallest integer such that $1<n \cdot a_1<p$ (it can be verified that such $n$ exists). Now, we have: $$\lfloor a \rfloor \cdot n < \lfloor a \cdot n \rfloor < \lfloor a \rfloor \cdot n + p$$so $\lfloor a \cdot n \rfloor$ cannot be a multiple of $p$, which is absurd (note that $\lfloor a \rfloor$ is a multiple of $p$). Similarly, it can't be $|q|>1$, so $|\frac{a}{b}|=1 \Longleftrightarrow |a|=|b|$. We can see that if $a=-b$, we have $-\lfloor a \cdot n \rfloor=\lfloor -a \cdot n \rfloor$, which for $n=1$ gives $-\lfloor a \rfloor=\lfloor -a \rfloor$. Since $a_1>0$, it is $\lfloor -a \rfloor=-\lfloor a \rfloor-1$, contradiction. If $a=b$, the condition is satisfied. So, the solutions from this case are the pairs of the form $(a, a)$, where $a$ is a real number which is not an integer. If WLOG $a$ is an integer different from 0, we have: $$\lfloor a \cdot n \rfloor=a \cdot n \Longrightarrow$$$$\lfloor b \cdot n \rfloor=b \cdot n \Longrightarrow$$$$\lfloor b \rfloor=b$$for $n=1$. So, $b$ is also an integer and we have the solutions $(a, b)$, where $a$, $b$ are arbitrary integers. If $a=0$ or $b=0$, the equality is true, so we have the solutions $(a, 0)$, $(0,a)$, where $a$ is a real number.

It's Problem N2 from the IMO Shortlist 1998 and has of course been posted before. See here.