In other words we want to find $$S=\sum_{m=1}^n m(-1)^{\tau(m)}$$It is well known that $\tau(m)$ is odd if and only if $m$ is a perfect square, so we infact have $$S=\sum_{m=1}^n m-2\sum_{k^2=m\leq n}m=$$$$\frac{n(n+1)}{2}-2\frac{\lfloor\sqrt n\rfloor(\lfloor\sqrt n\rfloor)}{2}=\frac{n^2+n}{2}-(\lfloor\sqrt n\rfloor^2+\lfloor\sqrt n\rfloor)$$

cadaeibf wrote: $$S=\sum_{m=1}^n m-2\sum_{k^2=m\leq n}m=$$$$\frac{n(n+1)}{2}-2\frac{\lfloor\sqrt n\rfloor(\lfloor\sqrt n\rfloor)}{2}=\frac{n^2+n}{2}-(\lfloor\sqrt n\rfloor^2+\lfloor\sqrt n\rfloor)$$ This is wrong. It should be $$S=\sum_{m=1}^n m-2\sum_{k^2=m\leq n}m=$$$$\frac{n(n+1)}{2}-2\cdot\frac{\lfloor\sqrt n\rfloor(\lfloor\sqrt n\rfloor+1)(2\lfloor\sqrt n\rfloor+1)}{6}=\frac{n^2+n}{2}-\frac{2\lfloor\sqrt n\rfloor^3+3\lfloor\sqrt n\rfloor^2+\lfloor\sqrt n\rfloor}{3}$$