I claim there are no solutions. If $a$ is integral, then we have $$\frac{n-1}{a} \leq \lfloor bn \rfloor < \frac{n}{a}.$$ Take $n = a,$ we see there are no integral integers between $\frac{n-1}{a}$ and $\frac{n}{a},$ so no solutions. If $a$ is a non-integral and bigger than $2$, by taking $n = \lfloor a \rfloor,$ we see than $$\frac{n-1}{a} = \frac{ \lfloor a \rfloor-1}{ \lfloor a \rfloor + \{a\}}<1$$ and $$\frac{n}{a} = \frac{ \lfloor a \rfloor}{ \lfloor a \rfloor + \{a\}} < 1$$ so there are no positive integer between these two, so no possible value of $\lfloor bn \rfloor.$ No possible $b$. If $1<a<2$ is a solution, let $a = 1+m, $ with $m<1.$ Notice by taking $n=1,$ we have $$\frac{n-1}{a} = 0 \leq \lfloor bn \rfloor < \frac{n}{a} = \frac{1}{1+m} < 1$$ so $ \lfloor bn \rfloor = 0$ so $b=0,$ which is not positive. (Sorry this step is wrong. It is 0<b<1. However, this is impossible.) Lastly, if $a = m,$ with $m<1,$ we have $$\frac{n-1}{m} \leq \lfloor bn \rfloor < \frac{n}{m}.$$ Now, suppose $\lfloor bn \rfloor = k,$ with $\frac{n-1}{m} \leq k < \frac{n}{m}.$ Therefore, $$\frac{k}{n} \leq b < \frac{k+1}{n}.$$ However, we have $$\lim_{k,n\to\infty} \frac{k}{n} = \lim_{k, n \to\infty} \frac{k+1}{n},$$ so no solutions for $b.$ We have exhausted all cases of $a$, so there are no pairs of real numbers $a, b$ such that $\lfloor a \lfloor bn \rfloor \rfloor =n - 1$ for all positive integers $n.$