Does Egyptian fraction help?

Suppose that $n>2017$ is a positive integer, and that $a,b,c$ are pairwise coprime positive integers such that $\frac{2017}{n} \; = \; \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$ Then $n \; = \; \frac{2017abc}{ab + ac + bc}$ If $p$ was a prime number dividing both $ab + ac + bc$ and $abc$ , then (since it divides $abc$ ) $p$ must divide one of $a,b,c$ . For definiteness, suppose that $p$ divides $a$ . Now, since $p$ divides $a$ and $ab+ac+bc$ , it follows that $p$ divides $bc$ , so divides one of $b$ and $c$ . Thus $a,b,c$ are not pairwise coprime. Thus we deduce that $abc$ and $ab +ac + bc$ are coprime. This implies that $ab + ac + bc$ divides $2017$ and hence either $ab+ac+bc$ equals either $1$ or $2017$ . It is not possible that $ab+ac+bc=1$ , and hence $ab+ac+bc=2017$ , and $n=abc$ .

Checking shows that there are only $6$ solutions in positive integers of the equation $ab + ac + bc = 2017$ with (for definiteness) $a \ge b \ge c$ , and these are $\begin{array}{|cccc|}\hline a & b & c & n=abc \\ \hline 39 & 22 & 19 & 16302 \\ 44 & 23 & 15 & 15180 \\ 45 & 41 & 2 & 3690 \\ 63 & 19 & 10 & 11970 \\ 103 & 15 & 4 & 6180 \\ 1008 & 1 & 1 & 1008 \\ \hline \end{array}$

To find these solutions, note that $2017 \ge 3c^2$ , and hence $c \le 25$ . For each $1 \le c \le 25$ we need to solve the equation $(a+c)(b+c) \; = \; 2017 + c^2$ and so $2017 + c^2$ must be writable as a product of two integers, both at least $2c$ . We just have to run through these $25$ cases to complete the above table.

The last solution in the table gives too small a value of $n$ , and so we see that $\boxed{3690}$ is the smallest integer for which the equations can be satisfied.