Let $a_1=-1$ and $a_n=-1-\sum_{j=1}^{n-1}a_j[n/j]$, where $[x]$ is the integer part of $x$. Is it always true that $a_n\in\{-1,0,1\}$?
These coefficients are determined by the formal equality $\chi_{(0,1)}(x)=\sum_{n=1}^\infty a_n\rho(1/(nx))$, where $\rho(x)=x-[x]$, which is related to the characterization of the Riemann Hypothesis in terms of the approximation in $L^2(0,\infty)$ of the above characteristic function by the linear combinations of the sequence $\{\rho(1/(nx))\}_n$
Let's identify your coefficients $a_j$: Your equation can be rewritten as $$\sum_{j=1}^n a_j\lfloor n/j\rfloor=-1,$$ since $\lfloor n/n\rfloor=1$ (I'll use the notation $\lfloor x\rfloor$ for the integer part of $x$, as it is more common nowadays than the $[x]$ used in my youth). For $n-1$ instead of $n$, this is $$\sum_{j=1}^{n-1} a_j\lfloor (n-1)/j\rfloor=-1,$$ but can also be written $$\sum_{j=1}^n a_j\lfloor (n-1)/j\rfloor=-1,$$ because $\lfloor (n-1)/n\rfloor=0$. Subtraction of both equations gives $$\sum_{j=1}^n a_j(\lfloor n/j\rfloor-\lfloor (n-1)/j\rfloor)=0.$$ Now it's easy to see that $\lfloor n/j\rfloor-\lfloor (n-1)/j\rfloor$ is $0$ if $n$ is not a multiple of $j$, and $1$ if it is. So our equation becomes $$\sum_{j|n}a_j=0.$$ This is exactly the equation satisfied by the Möbius function, $$\sum_{j|n}\mu(j)=0,$$ just with the starting value $a_1=-1$ instead of $\mu(1)=1$. This means we have $a_j=-\mu(j)$ for all $j\ge1$.