IMO Hk TST

The greatest possible value of $d(m+n)$ is $166635$.
Let $m = da$ and $n = db$ so that $(a, b) = 1$. The condition becomes $d \mid 3(a - b)$. WLOG assume $m > n$. Let $N = d(m + n) = d^2(a + b)$.
If $3 \nmid d$, then $d \mid a-b$. This implies $d \le a-b$. Also, we have $a = \frac{m}{d} \le \frac{2013}{d}$. This yields
$$N = d^2(a + b) \le d^2(2a - d) \le 4026d - d^3.$$

By differentiation, the maximum value of $4026d - d^3$ is attained at $d = \sqrt{\frac{4026}{3}}$.
Thus,
$$N \le \sqrt{\frac{4026}{3}} \left(4026 - \frac{4026}{3}\right) < 40 \cdot 3000 = 120000.$$
If $3 \mid d$, then $d \le 3(a-b)$. Also, we have $a \le \lfloor \frac{2013}{d} \rfloor$. This yields
$$N = d^2(a+b) \le d^2 \left( 2a - \frac{d}{3} \right) \le d^2 \left( 2 \lfloor \frac{2013}{d} \rfloor - \frac{d}{3} \right).$$
* For $d \le 54$, we have $N \le 4026d - \frac{d^3}{3} \le 4026(54) - \frac{(54)^3}{3} = 164916$ since $f(d) = 4026d - \frac{d^3}{3}$ is increasing for $d \le 54$ (again by differentiation).
* For $d = 57$, we have $N \le (57)^2 \left(2 \lfloor \frac{2013}{57} \rfloor - \frac{57}{3}\right) = 165699$.
* For $d = 60$, we have $N \le (60)^2 \left(2 \lfloor \frac{2013}{60} \rfloor - \frac{60}{3}\right) = 165600$.
* For $d = 63$, we have $N \le (63)^2 \left(2 \lfloor \frac{2013}{63} \rfloor - \frac{63}{3}\right) = 162729$.
* For $d = 66$, we have $N \le (66)^2 \left(2 \lfloor \frac{2013}{66} \rfloor - \frac{66}{3}\right) = 165528$.
* For $d = 69$, we have $N \le (69)^2 \left(2 \lfloor \frac{2013}{69} \rfloor - \frac{69}{3}\right) = 166635$.
* For $d \ge 72$, we have $N \le 4026d - \frac{d^3}{3} \le 4026(72) - \frac{(72)^3}{3} = 165456$ since $f(d)$ is decreasing for $d \ge 54$.
To conclude, we have $N \le 166635$. Equality holds when $d = 69$, $a = 29$ and $b = 6$. This means $(m, n) = (2001, 414)$.