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
$$a + b = 1000 - 707 = 293, \quad a + 2b = [f(707)] - [f(1000)] = 1414 - 1000 = 414.$$
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 $172+1000 = 1172$.