Find all positive integers $n < 589$ for which $589$ divides $n^2 + n + 1$.
To find all positive integers \( n < 589 \) for which \( 589 \) divides \( n^2 + n + 1 \), we start by factoring \( 589 \) into its prime factors:
\[ 589 = 19 \times 31 \]

We need \( 589 \mid n^2 + n + 1 \), which implies:
\[ n^2 + n + 1 \equiv 0 \pmod{19} \]
\[ n^2 + n + 1 \equiv 0 \pmod{31} \]

### Step 1: Solve modulo 19
First, consider the congruence modulo 19:
\[ n^2 + n + 1 \equiv 0 \pmod{19} \]

This can be rewritten as:
\[ n^2 + n + 1 \equiv 0 \pmod{19} \]
\[ 2n^2 + 2n + 2 \equiv 0 \pmod{19} \]
\[ (2n + 1)^2 \equiv 1 - 3 \pmod{19} \]
\[ (2n + 1)^2 \equiv -3 \pmod{19} \]
\[ (2n + 1)^2 \equiv 16 \pmod{19} \]

Taking the square root on both sides, we get:
\[ 2n + 1 \equiv \pm 4 \pmod{19} \]

This gives us two cases:
\[ 2n + 1 \equiv 4 \pmod{19} \]
\[ 2n \equiv 3 \pmod{19} \]
\[ n \equiv \frac{3}{2} \pmod{19} \]
\[ n \equiv 7 \pmod{19} \]

And:
\[ 2n + 1 \equiv -4 \pmod{19} \]
\[ 2n \equiv -5 \pmod{19} \]
\[ n \equiv \frac{-5}{2} \pmod{19} \]
\[ n \equiv 11 \pmod{19} \]

Thus, we have:
\[ n \equiv 7 \pmod{19} \]
\[ n \equiv 11 \pmod{19} \]

### Step 2: Solve modulo 31
Next, consider the congruence modulo 31:
\[ n^2 + n + 1 \equiv 0 \pmod{31} \]

This can be rewritten as:
\[ n^2 + n + 1 \equiv 0 \pmod{31} \]
\[ 2n^2 + 2n + 2 \equiv 0 \pmod{31} \]
\[ (2n + 1)^2 \equiv 1 - 3 \pmod{31} \]
\[ (2n + 1)^2 \equiv -3 \pmod{31} \]
\[ (2n + 1)^2 \equiv 121 \pmod{31} \]

Taking the square root on both sides, we get:
\[ 2n + 1 \equiv \pm 11 \pmod{31} \]

This gives us two cases:
\[ 2n + 1 \equiv 11 \pmod{31} \]
\[ 2n \equiv 10 \pmod{31} \]
\[ n \equiv \frac{10}{2} \pmod{31} \]
\[ n \equiv 5 \pmod{31} \]

And:
\[ 2n + 1 \equiv -11 \pmod{31} \]
\[ 2n \equiv -12 \pmod{31} \]
\[ n \equiv \frac{-12}{2} \pmod{31} \]
\[ n \equiv -6 \pmod{31} \]
\[ n \equiv 25 \pmod{31} \]

Thus, we have:
\[ n \equiv 5 \pmod{31} \]
\[ n \equiv 25 \pmod{31} \]

### Step 3: Combine solutions using the Chinese Remainder Theorem
We now combine the solutions using the Chinese Remainder Theorem.

For \( n \equiv 7 \pmod{19} \) and \( n \equiv 5 \pmod{31} \):
\[ n = 19k + 7 \]
\[ 19k + 7 \equiv 5 \pmod{31} \]
\[ 19k \equiv -2 \pmod{31} \]
\[ 19k \equiv 29 \pmod{31} \]
\[ k \equiv 29 \cdot 19^{-1} \pmod{31} \]

The inverse of 19 modulo 31 is 10 (since \( 19 \cdot 10 \equiv 1 \pmod{31} \)):
\[ k \equiv 29 \cdot 10 \pmod{31} \]
\[ k \equiv 290 \pmod{31} \]
\[ k \equiv 11 \pmod{31} \]

Thus:
\[ n = 19k + 7 \]
\[ n = 19(31m + 11) + 7 \]
\[ n = 589m + 216 \]

For \( n \equiv 7 \pmod{19} \) and \( n \equiv 25 \pmod{31} \):
\[ n = 19k + 7 \]
\[ 19k + 7 \equiv 25 \pmod{31} \]
\[ 19k \equiv 18 \pmod{31} \]
\[ k \equiv 18 \cdot 19^{-1} \pmod{31} \]
\[ k \equiv 18 \cdot 10 \pmod{31} \]
\[ k \equiv 180 \pmod{31} \]
\[ k \equiv 25 \pmod{31} \]

Thus:
\[ n = 19k + 7 \]
\[ n = 19(31m + 25) + 7 \]
\[ n = 589m + 482 \]

For \( n \equiv 11 \pmod{19} \) and \( n \equiv 5 \pmod{31} \):
\[ n = 19k + 11 \]
\[ 19k + 11 \equiv 5 \pmod{31} \]
\[ 19k \equiv -6 \pmod{31} \]
\[ 19k \equiv 25 \pmod{31} \]
\[ k \equiv 25 \cdot 19^{-1} \pmod{31} \]
\[ k \equiv 25 \cdot 10 \pmod{31} \]
\[ k \equiv 250 \pmod{31} \]
\[ k \equiv 2 \pmod{31} \]

Thus:
\[ n = 19k + 11 \]
\[ n = 19(31m + 2) + 11 \]
\[ n = 589m + 49 \]

For \( n \equiv 11 \pmod{19} \) and \( n \equiv 25 \pmod{31} \):
\[ n = 19k + 11 \]
\[ 19k + 11 \equiv 25 \pmod{31} \]
\[ 19k \equiv 14 \pmod{31} \]
\[ k \equiv 14 \cdot 19^{-1} \pmod{31} \]
\[ k \equiv 14 \cdot 10 \pmod{31} \]
\[ k \equiv 140 \pmod{31} \]
\[ k \equiv 16 \pmod{31} \]

Thus:
\[ n = 19k + 11 \]
\[ n = 19(31m + 16) + 11 \]
\[ n = 589m + 315 \]

### Step 4: List all solutions less than 589
We list all solutions \( n < 589 \):
\[ n = 49, 216, 315, 482 \]

The final answer is \( \boxed{49, 216, 315, 482} \)