Mongolian Mathematical Olympiad

$\frac{3}{2014} = \frac{1}{p} + \frac{1}{q} \Rightarrow 3pq = 2014(p+q) = 2 \cdot 19 \cdot 53(p+q).$

i) Consider the case $2 \cdot 19 \cdot 53 \mid p$. Setting $p = 2 \cdot 19 \cdot 53r$ we get $3rq = 2 \cdot 19 \cdot 53r + q \Rightarrow q = \frac{2 \cdot 19 \cdot 53r}{3r-1}$. Since $q \in \mathbb{N}$, $3r-1\mid 2 \cdot 19 \cdot 53r$, $(r, 3r-1) = 1 \Rightarrow 3r-1$ may take values 2, 38, 53, 1007. Therefore
$$\begin{align*} 3r - 1 &= 2 \Rightarrow & r &= 1, \\ 3r - 1 &= 38 \Rightarrow & r &= 13, \\ 3r - 1 &= 53 \Rightarrow & r &= 18, \\ 3r - 1 &= 1007 \Rightarrow & r &= 336 \end{align*}$$
and corresponding 4 pairs $(p, q)$ are $(2014, 1007)$, $(26182, 689)$, $(36252, 684)$, $(676704, 672)$.

ii) Consider the case $2 \cdot 19 \nmid p$ and $53 \nmid p$. Setting $p = 2 \cdot 19 \cdot r$ we get $3 \cdot 2 \cdot 19 \cdot rq = 2 \cdot 19 \cdot 53(2 \cdot 19r + q)$. Let $q = 53s$. Consequently, $s = \frac{2 \cdot 19r}{3r - 53}$, $(r, 3r - 53) = 1 \Rightarrow 3r - 53\mid 2 \cdot 19r$. Therefore
$$\begin{align*} 3r - 53 &= 1 \Rightarrow & r &= 18, \\ 3r - 53 &= 19 \Rightarrow & r &= 24 \end{align*}$$

iii) Consider the case $2 \cdot 53 \nmid p$ and $19 \nmid p$. From here follows $\Rightarrow p = 2 \cdot 53r$ and setting $q = 19s$ we get $s = \frac{2 \cdot 53r}{3r - 19}$. It implies $3r - 19 \mid 2 \cdot 53r$. $3r - 19 = 2 \Rightarrow r = 7$
$3r - 19 = 53 \Rightarrow r = 24.$
If $r = 7$ then $p = 2 \cdot 7 \cdot 53$, $q = 7 \cdot 19 \cdot 53 \Rightarrow (p, q) = (742, 7049)$
If $r = 24$ then $p = 2 \cdot 24 \cdot 53$, $q = 19 \cdot 48 \Rightarrow (p, q) = (2544, 912)$

iv) Consider the case $19 \cdot 53 \nmid p$ and $2 \nmid p$. Setting $p = 19 \cdot 53r$, $q = 2s$ we get $s = \frac{19 \cdot 53r}{3r-2}$. Therefore $3r - 2 \mid 19 \cdot 53r$.
$3r - 2 = 1 \Rightarrow r = 1;$
$3r-2=19 \Rightarrow r=7$. Corresponding pairs are $(p,q) = (1007, 2014); (7049, 742)$.

Thus there are 6 possibilities: $\frac{3}{2014} = \frac{1}{2014} + \frac{1}{1007} = \frac{1}{26182} + \frac{1}{689} =$
$$\frac{1}{676704} + \frac{1}{672} = \frac{1}{36252} + \frac{1}{684} = \frac{1}{912} + \frac{1}{2544} = \frac{1}{7049} + \frac{1}{742}.$$