Problem 1405

1. Rearrange the given equation:
$$\frac{1}{a-b} - \frac{1}{a} + \frac{1}{b} = \frac{1}{n}$$
To:
$$\frac{b(a-b) - a(a-b) + ab}{ab(a-b)} = \frac{1}{n}$$
Simplifying the numerator:
$$b(a-b) - a(a-b) + ab = ab - b^2 - a^2 + ab + ab = 3ab - a^2 - b^2$$
Thus, the equation becomes:
$$\frac{3ab - a^2 - b^2}{ab(a-b)} = \frac{1}{n}$$
Rearranging gives:
$$n(3ab - a^2 - b^2) = ab(a-b)$$
Simplifying further:
$$n(a^2 - ab + b^2) = ab(a-b)$$

2. **Set $a = da_1$ and $b = db_1$ where $d = \gcd(a, b)$:**
$$n(d^2(a_1^2 - a_1b_1 + b_1^2)) = d^2a_1b_1(a_1 - b_1)$$
Dividing both sides by $d^2$:
$$n(a_1^2 - a_1b_1 + b_1^2) = a_1b_1(a_1 - b_1)$$

3. Analyze the gcd condition:
$$\gcd(a_1^2 - a_1b_1 + b_1^2, a_1b_1) = 1$$
If $p$ is a prime dividing both $a_1^2 - a_1b_1 + b_1^2$ and $a_1 - b_1$, then:
$$a_1 \equiv b_1 \pmod{p}$$
Substituting $a_1 \equiv b_1 \pmod{p}$ into $a_1^2 - a_1b_1 + b_1^2$:
$$a_1^2 - a_1^2 + a_1^2 \equiv 0 \pmod{p}$$
Thus, $p \mid a_1, b_1$, which is a contradiction.

4. **Find $\ell$ such that:**
$$d = \ell(a_1^2 - a_1b_1 + b_1^2) \quad \text{and} \quad n = \ell a_1b_1(a_1 - b_1)$$

5. **Consider primes $p_1 < p_2 < \cdots < p_L$ where $L \gg 2024$:**
$$n = \prod_{1 \le i \le L} p_i(p_i - 1)$$
For $a_1 = p_i$ and $b_1 = 1$:
$$\ell = \frac{\prod_j p_j(p_j - 1)}{p_i(p_i - 1)}$$
This generates at least $L$ distinct pairs $(a, b)$.

6. Conclusion:
Since $L$ can be chosen to be greater than 2024, there exists an integer $n \geq 1$ such that the number of pairs $(a, b)$ satisfying the given equation exceeds 2024.

$\blacksquare$