Proving $\sum_{1}^{n} \left\lceil\log_{2}\frac{2n}{2i-1}\right\rceil=2n -1 $ 
Show that $$\sum_{i=1}^{n} \left\lceil\log_{2}\frac{2n}{2i-1}\right\rceil=2n -1 $$ where $ \lceil\cdot\rceil$ denotes the ceiling function.

My method: one way would be observe each part of the summation is telling numbers of form $2^\alpha k$ where $k$ is any odd number (from the list of numbers from $2$ to $2n$).
But I would like to know whether there is an algebraic proof, or maybe some bit of number theory that will help solve it.
 A: You can prove it by induction. The first observation is:
$$\left\lceil\log_{2}\frac{2n}{2i-1}\right\rceil = 1 + \left\lceil\log_{2}\frac{n}{2i-1}\right\rceil$$
So it suffices to prove $\sum_{i=1}^{n} \left\lceil\log_{2}\frac{n}{2i-1}\right\rceil = n-1$. The next observation is that the latter half of the sum is zero (when $2i - 1 > n$, we have $\frac{n}{2i-1} \in (\frac 12, 1)$). Formally:
$n=2k$:
$$\sum_{i=1}^{2k} \left\lceil\log_{2}\frac{2k}{2i-1}\right\rceil = \sum_{i=1}^{k} \left\lceil\log_{2}\frac{2k}{2i-1}\right\rceil + 0,$$
which by induction hypothesis is $2k-1 = n - 1$.
$n=2k + 1$:
$$\sum_{i=1}^{2k} \left\lceil\log_{2}\frac{2k + 1}{2i-1}\right\rceil = \sum_{i=1}^{k + 1} \left\lceil\log_{2}\frac{2k + 1}{2i-1}\right\rceil + 0 = 
1 + \sum_{i=1}^{k} \left\lceil\log_{2}\frac{2k + 1}{2i-1}\right\rceil.$$
We want to show that $\sum_{i=1}^{k} \left\lceil\log_{2}\frac{2k + 1}{2i-1}\right\rceil = (n - 1) - 1 = 2k$.
By the induction hypothesis, we know that $\sum_{i=1}^{k} \left\lceil\log_{2}\frac{2k}{2i-1}\right\rceil = 2k - 1$.
Let's compare them term by term, i.e. $\left\lceil\log_{2}\frac{2k}{2i-1}\right\rceil$ vs $\left\lceil\log_{2}\frac{2k + 1}{2i-1}\right\rceil$.
Let's fix $i$, and let $\ell$ be the smallest integer such that $2^\ell \ge \frac{2k}{2i-1}$.
When is it not the case that $2^\ell \ge \frac{2k + 1}{2i - 1}$? It happens iff $\frac{2k}{2i - 1}$ is a power of two.
This happens for exactly on term: for $i$ such that $2k = (2i - 1) \cdot 2^t$ for some $t$.
Hence,
$$\sum_{i=1}^{k} \left\lceil\log_{2}\frac{2k + 1}{2i-1}\right\rceil = 1 +\sum_{i=1}^k \left\lceil\log_{2}\frac{2k}{2i-1}\right\rceil = 2k$$
