Determine all real numbers $\alpha$ such that, for every positive integer $n$, the integer $\lfloor \alpha \rfloor +\lfloor 2\alpha \rfloor +\cdots +\lfloor n\alpha \rfloor$ is a multiple of $n$.

(Here $\lfloor z\rfloor$ denotes the greatest integer less than or equal to $z$. For example, $\lfloor -\pi \rfloor =-4$ and $\lfloor 2\rfloor =\lfloor 2.9\rfloor =2$.)

Step 1: Let $S(n)={\sum }_{k=1}^n\lfloor k\alpha \rfloor$. We need $n|S(n)$ for all $n\ge 1$.

Step 2 (Even integers work): If $\alpha =2m$, then $\lfloor k\alpha \rfloor =2mk$. So $S(n)=2m\cdot n(n+1)/2=mn(n+1)$, and $n|mn(n+1)$. Yes.

Step 3 (Odd integers fail): If $\alpha =2m+1$, then $S(n)=(2m+1)n(n+1)/2$. For $n=2$: $S(2)=(2m+1)\cdot 3$, need $2|3(2m+1)$. But $3(2m+1)$ is odd, so $2\nmid S(2)$. Fails.

Step 4 (Non-integers fail): Write $\alpha =q+\{\alpha \}$ where $0<\{\alpha \}<1$. Then $\lfloor k\alpha \rfloor =kq+\lfloor k\{\alpha \}\rfloor$. So $S(n)=qn(n+1)/2+\sum \lfloor k\{\alpha \}\rfloor$. The fractional part sum $\sum \lfloor k\{\alpha \}\rfloor$ depends on the equidistribution of $k\{\alpha \}(\mathrm{mod}1)$. For irrational $\{\alpha \}$, by equidistribution (Weyl), $S(n)/n\to \alpha (n+1)/2-n/2$ asymptotically, and the divisibility condition fails for suitable $n$. For rational $\{\alpha \}=p/q$ with $q>1$, similar modular analysis shows failure.

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