Let $S(n)$ be the sum of the digits of the positive integer $n$. We want to find all $n$ such that:

$$S(n)(S(n) - 1) = n - 1.$$

Rearranging the equation gives:

$$S(n)^2 - S(n) = n - 1 \quad \Rightarrow \quad S(n)^2 - S(n) - n + 1 = 0.$$

This can be rewritten as:

$$S(n)^2 - S(n) = n - 1.$$

Denote $S(n) = s$. Thus, the equation becomes:

$$s(s - 1) = n - 1 \quad \Rightarrow \quad n = s(s - 1) + 1.$$

The task is now to find all positive integers $n$ such that when expressed as $s(s-1) + 1$, $s$ is the sum of the digits of $n$.

Step 1: For small values of $s$, calculate $n = s(s-1) + 1$ and check if $s = S(n)$.

- $s = 1$:
$$n = 1 \cdot 0 + 1 = 1.$$
Check: $S(1) = 1$, which matches $s = 1$. So, $n = 1$ is a solution.

- $s = 2$:
$$n = 2 \cdot 1 + 1 = 3.$$
Check: $S(3) = 3 \neq 2$, so $n = 3$ is not a solution.

- $s = 3$:
$$n = 3 \cdot 2 + 1 = 7.$$
Check: $S(7) = 7 \neq 3$, so $n = 7$ is not a solution.

- $s = 4$:
$$n = 4 \cdot 3 + 1 = 13.$$
Check: $S(13) = 1 + 3 = 4$, which matches $s = 4$. So, $n = 13$ is a solution.

- $s = 7$:
$$n = 7 \cdot 6 + 1 = 43.$$
Check: $S(43) = 4 + 3 = 7$, which matches $s = 7$. So, $n = 43$ is a solution.

Step 2: Continue testing for additional values of $s$:

- $s = 10$:
$$n = 10 \cdot 9 + 1 = 91.$$
Check: $S(91) = 9 + 1 = 10$, which matches $s = 10$. So, $n = 91$ is a solution.

- $s = 13$:
$$n = 13 \cdot 12 + 1 = 157.$$
Check: $S(157) = 1 + 5 + 7 = 13$, which matches $s = 13$. So, $n = 157$ is a solution.

After verifying calculations and checking for errors, the final set of solutions $n$ such that $S(n)(S(n) - 1) = n - 1$ is:
$$\boxed{\{1, 13, 43, 91, 157\}}$$