Find all positive integers $a, b, c$ such that $1/a + 1/b + 1/c = 4/3999$.
The contest is just ended, so you may freely answer. (I did not attend the contest: it is an Italian contest for schools and the groups may ask other people for hints. This is how I got involved, but of course asking on MSE would have been cheating)
I found three solutions: I am explaining below what I did. I would like however to know if there is a more general approach.
If we have more general equation $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{n}{d},$$ where $a,b,c,d,n\in \mathbb{N}$, $a\leqslant b \leqslant c$, ($n<d$), then search algorithm can be described as:
1) since $\displaystyle\frac{1}{a}$ is the greater, then $\displaystyle\frac{1}{a}\geqslant\frac{1}{3}\cdot \frac{n}{d}$.
So, we'll consider every $a\in\mathbb{N}$, such that
$$\displaystyle\frac{d}{n} < a \leqslant \frac{3d}{n}.$$
2) since $\displaystyle\frac{1}{b}$ is the 2-nd greater, then $\displaystyle\frac{1}{b}\geqslant\frac{1}{2}\cdot \left(\frac{n}{d}-\frac{1}{a}\right)$.
So, we'll consider every $b\in\mathbb{N}$, $b\geqslant a$, such that $\displaystyle\left(\frac{n}{d}-\frac{1}{a}\right)^{-1}< b \leqslant 2 \cdot \left(\frac{n}{d}-\frac{1}{a}\right)^{-1}$, $$\frac{ad}{an-d} < b \leqslant \frac{2ad}{an-d} .$$
3) when we have $a,b$, we consider fraction $$\frac{n}{d}-\frac{1}{a}-\frac{1}{b},$$ and calculate 2 values (numerator and denominator): $$N = abn - bd - ad,$$ $$D = abd.$$
If $N$ divides $D$, then $c=D/N$ is integer, $c\geqslant b$, and we can add triple $(a,b,c)$ to solution list.
Using this algorithm, I found 1834 ordered solutions:
1) $\displaystyle\frac{1}{1000} + \frac{1}{3999001} + \frac{1}{15992004999000}$;
2) $\displaystyle\frac{1}{1000}+\frac{1}{3999002}+\frac{1}{7996004499000}$;
3) $\displaystyle\frac{1}{1000}+\frac{1}{3999003}+\frac{1}{5330670999000}$;
4) $\displaystyle\frac{1}{1000}+\frac{1}{3999004}+\frac{1}{3998004249000}$;
5) $\displaystyle\frac{1}{1000}+\frac{1}{3999005}+\frac{1}{3198404199000}$;
6) $\displaystyle\frac{1}{1000}+\frac{1}{3999006}+\frac{1}{2665337499000}$;
7) $\displaystyle\frac{1}{1000}+\frac{1}{3999008}+\frac{1}{1999004124000}$;
...
1832) $\displaystyle\frac{1}{2666}+\frac{1}{2666}+\frac{1}{3999}$;
1833) $\displaystyle\frac{1}{2666}+\frac{1}{2709}+\frac{1}{3906}$;
1834) $\displaystyle\frac{1}{2924}+\frac{1}{2924}+\frac{1}{3162}$.
Among them (with small $a,b,c$-values) we can find:
$\displaystyle\frac{1}{1806}+\frac{1}{4123}+\frac{1}{4902}$,
$\displaystyle\frac{1}{2046}+\frac{1}{3311}+\frac{1}{4774}$,
$\displaystyle\frac{1}{2193}+\frac{1}{3162}+\frac{1}{4386}$,
$\displaystyle\frac{1}{2294}+\frac{1}{3182}+\frac{1}{3999}$,
$\displaystyle\frac{1}{2618}+\frac{1}{3162}+\frac{1}{3311}$.
More complete list: