Japan Mathematical Olympiad

First, we prove the following Lemma.

Lemma. If real numbers $x,y$ and an integer $k$ satisfy $k<x-y<k+1$, then $[x]-[y]=k$ or $k+1$ must hold. Proof: Since $0\le x-[x]<1$ and $0\le y-[y]<1$, we have $x-y-1<[x]-[y]<x-y+1$. This, together with $k<x-y<k+1$, yields $k-1<[x]-[y]<k+2$. Since $[x]-[y]$ is an integer, we obtain the assertion of the Lemma. For a positive integer $n$, let us write $f(n)=\frac{1000000}{n}$. Then, we get $f(n)-f(n+1)=\frac{1000000}{n(n+1)}$.

(1) Case where $1\le n<707$: We have $n(n+1)<500000$ so that $2<f(n)-f(n+1)$ holds in this case. Therefore, by the Lemma, we get $[f(n)]-[f(n+1)]\ge 2$, which shows that there are no $n$ satisfying the condition of the problem in this case.

(2) Case where $707\le n<1000$: We have $500000<n(n+1)<1000000$ and therefore, $1<f(n)-f(n+1)<2$ in this case. So, we have $[f(n)]-[f(n+1)]=1$ or $2$ by the Lemma. If we let $a$ ($b$) be the number of $n$'s with $707\le n<1000$, for which $[f(n)]-[f(n+1)]=1$ ($[f(n)]-[f(n+1)]=2$, respectively), then we have $$a+b=1000-707=293,a+2b=[f(707)]-[f(1000)]=1414-1000=414.$$ Solving these simultaneous equations, we obtain $a=172$.

(3) Case where $n\ge 1000$: We have $1000000<n(n+1)$, so that $0<f(n)-f(n+1)<1$ hold in this case. By the Lemma, we have $[f(n)]-[f(n+1)]=0$ or $1$. Since we have $[f(n)]=0$ if $n>1000$, the number of $n$'s for which $[f(n)]-[f(n+1)]=1$ in this case equals $[f(1000)]=1000$.

Therefore, we conclude that the desired answer for the problem is $172+1000=1172$.