1. We start with the given expression:
$$\sum_{k=1}^n \frac{a_k - b_k}{k}$$
where $a_k = \frac{1}{\binom{n}{k}}$ and $b_k = 2^{k-n}$.

2. We split the sum into two separate sums:
$$\sum_{k=1}^n \frac{a_k - b_k}{k} = \sum_{k=1}^n \frac{a_k}{k} - \sum_{k=1}^n \frac{b_k}{k}$$

3. Consider the first sum:
$$\sum_{k=1}^n \frac{a_k}{k} = \sum_{k=1}^n \frac{1}{k \binom{n}{k}}$$
Using the combinatorial identity $k \binom{n}{k} = n \binom{n-1}{k-1}$, we can rewrite the sum as:
$$\sum_{k=1}^n \frac{1}{k \binom{n}{k}} = \sum_{k=1}^n \frac{1}{n \binom{n-1}{k-1}}$$
Factoring out the constant $\frac{1}{n}$:
$$\sum_{k=1}^n \frac{1}{n \binom{n-1}{k-1}} = \frac{1}{n} \sum_{k=1}^n \frac{1}{\binom{n-1}{k-1}}$$

4. Recognize that the sum $\sum_{k=1}^n \frac{1}{\binom{n-1}{k-1}}$ is the sum of the reciprocals of the binomial coefficients for $n-1$. By the binomial theorem, we know:
$$\sum_{k=0}^{n-1} \binom{n-1}{k} = 2^{n-1}$$
Taking the reciprocals and summing, we get:
$$\sum_{k=1}^n \frac{1}{\binom{n-1}{k-1}} = 2^{n-1}$$
Therefore:
$$\frac{1}{n} \sum_{k=1}^n \frac{1}{\binom{n-1}{k-1}} = \frac{1}{n} \cdot 2^{n-1} = \frac{1}{2^{n-1}}$$

5. Now consider the second sum:
$$\sum_{k=1}^n \frac{b_k}{k} = \sum_{k=1}^n \frac{2^{k-n}}{k}$$
This can be rewritten as:
$$\sum_{k=1}^n \frac{2^{k-n}}{k} = \frac{1}{2^n} \sum_{k=1}^n \frac{2^k}{k}$$
Recognize that this is a geometric series:
$$\sum_{k=1}^n 2^k = 2 + 2^2 + 2^3 + \cdots + 2^n = 2 (1 + 2 + 2^2 + \cdots + 2^{n-1}) = 2 (2^n - 1) = 2^{n+1} - 2$$
Dividing by $2^n$:
$$\frac{1}{2^n} \sum_{k=1}^n 2^k = \frac{1}{2^n} (2^{n+1} - 2) = 2 - \frac{2}{2^n} = 2 - \frac{1}{2^{n-1}}$$

6. Combining the results from steps 4 and 5:
$$\sum_{k=1}^n \frac{a_k - b_k}{k} = \frac{1}{2^{n-1}} - \left(2 - \frac{1}{2^{n-1}}\right) = \frac{1}{2^{n-1}} - 2 + \frac{1}{2^{n-1}} = 0$$

Thus, we have shown that:
$$\sum_{k=1}^n \frac{a_k - b_k}{k} = 0$$

$\blacksquare$