<s>[INST] INMO Problem with even function proof. 
Let $n$ be a natural number. Show that
  $$\left[ \frac{n}{1} \right ] + \left[ \frac{n}{2} \right ] + \left[ \frac{n}{3} \right ] + \cdots + \left[ \frac{n}{n} \right ] + [\sqrt{n}]$$ is even. Here $[x]$ means the greatest integer less than or equal to $x$. 

Let $$f(n) = [\sqrt{n}] + \sum_{k=1}^{n} \left[ \frac{n}{k} \right]$$
$$f(1) = [1] + [1] = 2 \checkmark$$
The statement for $f(n+1)$ is to be proved:
$$f(n+1) = [\sqrt{n+1}] + \sum_{k=1}^{n+1} \left[\frac{n+1}{k} \right]$$
I just need some hints from now?
 [/INST] Hint: Prove by induction.
When is $ \lfloor \frac{n}{k} \rfloor \neq \lfloor \frac{n+1}{k} \rfloor $?
When they are not equal, what is the difference of these 2 terms?
Ans: The terms are different if there exists some integer $i$ such that $ \frac{n}{k} < i \leq \frac{ n+1}{k}$. Multiplying throughout by $k$, we get $ n < ik \leq n+1$. Since $n$, $n+1$ are consecutive integers, this then implies that $ik = n+1$, or that $k$ is a divisor of $n+1$.
Thus, this term increases by 1 if $k \mid n+1$. Otherwise, it stays constant.

Hence, letting $ g(n) = \sum_{i=1}^n \lfloor \frac{n}{k} \rfloor$, we get that   
$$g(n+1) - g(n) = \text{ number of divisors of } n$$
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