Solution:

We are to find all positive integers $n$ such that $5n - 4$ divides $n^2 - n + 1$.

Let $d = 5n - 4$. We require $d \mid n^2 - n + 1$.

This means there exists an integer $k$ such that:
$$n^2 - n + 1 = k(5n - 4)$$
Rewriting:
$$n^2 - n + 1 - 5k n + 4k = 0$$
$$n^2 - (5k + 1)n + (4k + 1) = 0$$
This is a quadratic in $n$. For $n$ to be integer, the discriminant must be a perfect square:
$$\Delta = [-(5k + 1)]^2 - 4 \cdot 1 \cdot (4k + 1) = (5k + 1)^2 - 4(4k + 1)$$
$$= 25k^2 + 10k + 1 - 16k - 4$$
$$= 25k^2 - 6k - 3$$
We want $\Delta$ to be a perfect square, say $m^2$:
$$25k^2 - 6k - 3 = m^2$$
So,
$$25k^2 - 6k - (m^2 + 3) = 0$$
We can try small integer values for $k$ to see if $\Delta$ is a perfect square.

Try $k = 0$:
$$\Delta = 0 - 0 - 3 = -3$$
Not a perfect square.

$k = 1$:
$$25 - 6 - 3 = 16$$
$16$ is a perfect square ($4^2$).

So $k = 1$ works. Let's find $n$:

Recall:
$$n^2 - (5k + 1)n + (4k + 1) = 0$$
For $k = 1$:
$$n^2 - 6n + 5 = 0$$
$$n = \frac{6 \pm \sqrt{36 - 20}}{2} = \frac{6 \pm 4}{2}$$
So $n = 5$ or $n = 1$.

Check if $n = 1$ and $n = 5$ work in the original condition:

For $n = 1$:
$5n - 4 = 1$, $n^2 - n + 1 = 1$; $1$ divides $1$.

For $n = 5$:
$5n - 4 = 21$, $n^2 - n + 1 = 25 - 5 + 1 = 21$; $21$ divides $21$.

Try $k = -1$:
$$25 - (-6) - 3 = 25 + 6 - 3 = 28$$
Not a perfect square.

Try $k = 2$:
$$100 - 12 - 3 = 85$$
Not a perfect square.

Try $k = -2$:
$$100 + 12 - 3 = 109$$
Not a perfect square.

Try $k = 3$:
$$225 - 18 - 3 = 204$$
Not a perfect square.

Try $k = -3$:
$$225 + 18 - 3 = 240$$
Not a perfect square.

Try $k = 4$:
$$400 - 24 - 3 = 373$$
Not a perfect square.

Try $k = -4$:
$$400 + 24 - 3 = 421$$
Not a perfect square.

Thus, the only positive integer solutions are $n = 1$ and $n = 5$.

Answer:

All positive integers $n$ such that $n^2 - n + 1$ is a multiple of $5n - 4$ are $n = 1$ and $n = 5$.