Interesting. Notice that $\left \lfloor {\frac{n+2^k}{2^{k+1}}}\right \rfloor=\left \lfloor {\frac{n}{2^k}}\right\rfloor -\left \lfloor \frac{n}{2^{k+1}}\right \rfloor$

I want to know if there is any other solution than the one given by Hawk Tiger!!!!!

Here is a rather contrived proof (but still a proof nonetheless): Consider the binary representation of $ n$: $ n=2^{u_0}+2^{u_1}+\cdots+2^{u_t}$. Then it is easy to see that $ \left\lfloor\frac{n+2^{k-1}}{2^k}\right\rfloor=2^{u_i-k}+2^{u_{i+1}-k}+\cdots+2^{u_t-k} +(0)\text{ if }u_{i-1}<k-1\text{ and }+(1)\text{ if }u_{i-1}=k-1$ for $ u_i\ge k$. Thus, the sum in question is equivalent to $ (2^{u_0-1}+2^{u_0-2}+\cdots+1)+(2^{u_1-1}+2^{u_1-2}+\cdots+1)+\cdots+(2^{u_t-1}+2^{u_t-2}+\cdots+1)+t+1$ $ =2^{u_0}-1+2^{u_1}-1+\cdots+2^{u_t}-1+t+1$ $ =n$

I am not sure but can someone (maybe peter) check out if this problem appeared in the 1968 IMO?

See here. The second solution is essentially Hawk Tiger's and the first is essentially the cosinator's.

Hint

and then represent $n$ as $n=2^{n} +m$ so that m is less than $2^{n}$.

Here's a way to see why the identity is true and also why Hawk Tiger's identity is true. Consider all the numbers less than or equal to $n$. We count which of these have $v_2$ equal to 0, 1, 2, and so on. Let $k$ be fixed;we are counting the number of nonnegative $m$ such that $2^{k+1}m + 2^k \le n$. This is simply $\left\lfloor\dfrac{n+2^k}{2^{k+1}}\right\rfloor$. Summing over all $k$, we get the desired result. For Hawk Tiger's identity, note that the first term is just the number of multiples of $2^k$ and we subtract the second term which is the number of multiples of $2^{k+1}$

awesome trick