Solution:

(Ⅰ) From the given: $a_{n+1} = \frac{2a_n}{a_n + 1}$,

∴ $\frac{1}{a_{n+1}} = \frac{a_n + 1}{2a_n} = \frac{1}{2} + \frac{1}{2} \cdot \frac{1}{a_n}$,

∴ $\frac{1}{a_{n+1}} - 1 = \frac{1}{2}\left(\frac{1}{a_n} - 1\right)$,

Given $a_1 = \frac{2}{3}$, ∴ $\frac{1}{a_1} - 1 = \frac{1}{2}$,

∴ The sequence $\left\{\frac{1}{a_n} - 1\right\}$ is a geometric sequence with the first term $\frac{1}{2}$ and common ratio $\frac{1}{2}$.

(Ⅱ) From (Ⅰ), we know $\frac{1}{a_n} - 1 = \frac{1}{2} \cdot \left(\frac{1}{2}\right)^{n-1} = \frac{1}{2^n}$,

Thus, $\frac{1}{a_n} = \frac{1}{2^n} + 1$, ∴ $\frac{n}{a_n} = \frac{n}{2^n} + n$.

Let $T_n = \frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \ldots + \frac{n}{2^n}$,

Then $\frac{1}{2}T_n = \frac{1}{2^2} + \frac{2}{2^3} + \ldots + \frac{n-1}{2^n} + \frac{n}{2^{n+1}}$,

From subtracting the above, we get: $\frac{1}{2}T_n = \frac{1}{2} + \frac{1}{2^2} + \ldots + \frac{1}{2^n} - \frac{n}{2^{n+1}} = \frac{\frac{1}{2}(1 - \frac{1}{2^n})}{1 - \frac{1}{2}} - \frac{n}{2^{n+1}} = 1 - \frac{1}{2^n} - \frac{n}{2^{n+1}}$,

∴ $T_n = 2 - \frac{1}{2^{n-1}} - \frac{n}{2^n}$. Also, $1 + 2 + 3 + \ldots + n = \frac{n(n+1)}{2}$.

∴ The sum of the first $n$ terms of the sequence $\left\{\frac{n}{a_n}\right\}$ is: $S_n = 2 - \frac{2+n}{2^n} + \frac{n(n+1)}{2} = \frac{n^2 + n + 4}{2} - \frac{2+n}{2^n}$.

Thus, the final answer is $\boxed{S_n = \frac{n^2 + n + 4}{2} - \frac{2+n}{2^n}}$.