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.
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, $$\sum_{i=1}^{k} \left\lceil\log_{2}\frac{2k + 1}{2i-1}\right\rceil = 1 +\sum_{i=1}^k \left\lceil\log_{2}\frac{2k}{2i-1}\right\rceil = 2k$$