Question 209, Find the number of positive integers $n$ that satisfy $\left[\frac{2012}{n}\right]-\left[\frac{2012}{n+1}\right]=1$, where $[x]$ denotes the integer part of $x$.
Question 209, Solution: According to the condition, we know that $1=\left[\frac{2012}{n}\right]-\left[\frac{2012}{n+1}\right]>\left(\frac{2012}{n}-1\right)-\frac{2012}{n+1}$, solving this yields $n \geq 32$. When $n \geq 2013$, $\left[\frac{2012}{n}\right]-\left[\frac{2012}{n+1}\right]=0$. Therefore, $32 \leq n \leq 2012$.
(1) When $45 \leq n \leq 2012$, $\frac{2012}{n} \leq \frac{2012}{n+1}+1$, thus $\left[\frac{2012}{n}\right] \leq\left[\frac{2012}{n+1}+1\right]=\left[\frac{2012}{n+1}\right]+1$. Therefore, when $45 \leq n \leq 2012$, $\left[\frac{2012}{n}\right]$ can take all values in $\{44,43, \ldots, 1\}$. For any $k \in \{44,43, \ldots, 1\}$, let the maximum positive integer solution of $\left[\frac{2012}{n}\right]=k$ be $a_{k}$, and the minimum positive integer solution be $b_{k}$, then $a_{k}+1=b_{k+1}$. Thus, when $b_{k} \leq n \leq a_{k}-1$, $\left[\frac{2012}{n}\right]-\left[\frac{2012}{n+1}\right]=0$; when $n=a_{k}$, $\left[\frac{2012}{n}\right]-\left[\frac{2012}{n+1}\right]=1$. Since $k$ has 44 values, there are exactly 44 $n \in \{45,46, \ldots, 2012\}$ such that $\left[\frac{2012}{n}\right]^{n}-\left[\frac{2012}{n+1}\right]^{n+1}=1$.
(2) For $32 \leq n \leq 44$, since $\frac{2012}{n+1}+1 \leq \frac{2012}{n} \leq \frac{2012}{n+1}+2$, hence $\left[\frac{2012}{n+1}\right]+1 \leq\left[\frac{2012}{n}\right] \leq \left[\frac{2012}{n+1}\right]+2$, thus $\left[\frac{2012}{n}\right]-\left[\frac{2012}{n+1}\right]=1$ or 2. Also, $\left[\frac{2012}{n}\right] \in A=\{45,46, \ldots, 62\}$. Note that $A$ has 18 elements, while $n$ has only 13 values, so there are exactly $18-13=5$ values of $n$ such that $\left[\frac{2012}{n}\right]-\left[\frac{2012}{n+1}\right]=2$, and thus $13-5=8$ values of $n$ such that $\left[\frac{2012}{n}\right]-\left[\frac{2012}{n+1}\right]=1$.
In summary, there are $44+8=52$ values of $n$ that satisfy the condition.