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$\phantom{=\quad}\sqrt{18\cdot n\cdot34}\\~\\ {=\quad}\sqrt{3\cdot3\cdot2\cdot n\cdot2\cdot17}\\~\\ {=\quad}\sqrt{3\cdot3\cdot2\cdot 2\cdot17\cdot n}\\~\\ {=\quad}\sqrt{3\cdot3}\cdot\sqrt{2\cdot 2}\cdot\sqrt{17\cdot n}\\~\\ {=\quad}3\cdot2\cdot\sqrt{17\cdot n}\\~\\ {=\quad}6\sqrt{17n}$

There must be a pair of  17's  in the prime factorization of the number under the sqrt.

So  17  must be a factor of  n  , and the least possible positive integer value of  n  is  17

$\phantom{=\quad}6\sqrt{17\cdot17}\\~\\ {=\quad}6\cdot17\\~\\ {=\quad}102$_