How many $m$ such that : $\sum\limits_{k=1}^m \left\lfloor\frac{m}{k}\right\rfloor$ be even? Find how many $m \le 1000$ such that : 
$$\sum\limits_{k=1}^m \left\lfloor \frac{m}{k}\right\rfloor$$
be even ( $\lfloor x\rfloor$ is the largest integer smaller than $x$.)
I think that one case is  $\sum\limits_{k=1}^m \left\lfloor \frac{m}{k}\right\rfloor$ is even if for every integer  $k$ ( $ 1 \le k \le m $) $\left\lfloor\frac{m}{k}\right\rfloor$ be even or the number of odd number be even  in $\sum\limits_{k=1}^m \left\lfloor\frac{m}{k}\right\rfloor$. $\big($ $
\sum\limits_{i=1}^n\left\lfloor\frac{n}{i}\right\rfloor=\sum\limits_{i=1}^n d(i)
$
where $d(\ )$ is the divisor function $\big)$

Hint, further to the comments
$$\sum\limits_{k=1}^{m} \left \lfloor \frac{m}{k} \right \rfloor=\sum\limits_{k=1}^{m}d(k) \tag{1}$$


*

*$m=1 \Rightarrow \sum\limits_{k=1}^{1} \left \lfloor \frac{1}{k} \right \rfloor=\sum\limits_{k=1}^{1}d(k)=\color{blue}{1}$, i.e. $\color{red}{\{1\}}$ is the only perfect square or $\color{blue}{\left \lfloor \sqrt{1} \right \rfloor=1}$ in total.

*$m=2 \Rightarrow \sum\limits_{k=1}^{2} \left \lfloor \frac{2}{k} \right \rfloor=\sum\limits_{k=1}^{2}d(k)=1+2=\color{blue}{3}$, i.e. $\color{red}{\{1\}}$ is the only perfect square or $\color{blue}{\left \lfloor \sqrt{2} \right \rfloor=1}$ in total.

*$m=3 \Rightarrow \sum\limits_{k=1}^{3} \left \lfloor \frac{3}{k} \right \rfloor=\sum\limits_{k=1}^{3}d(k)=1+2+2=\color{blue}{5}$, i.e. $\color{red}{\{1\}}$ is the only perfect square or $\color{blue}{\left \lfloor \sqrt{3} \right \rfloor=1}$ in total.

*$m=4 \Rightarrow \sum\limits_{k=1}^{4} \left \lfloor \frac{4}{k} \right \rfloor=\sum\limits_{k=1}^{4}d(k)=1+2+2+3=\color{blue}{8}$, i.e. $\color{red}{\{1, 4\}}$ are the only perfect squares or $\color{blue}{\left \lfloor \sqrt{4} \right \rfloor=2}$ in total.

*$m=5 \Rightarrow \sum\limits_{k=1}^{5} \left \lfloor \frac{5}{k} \right \rfloor=\sum\limits_{k=1}^{5}d(k)=1+2+2+3+2=\color{blue}{10}$, i.e. $\color{red}{\{1, 4\}}$ are the only perfect squares or $\color{blue}{\left \lfloor \sqrt{5} \right \rfloor=2}$ in total.


Is the pattern visible now? It should be easy to show it by induction.
