jmc

By the Euclidean algorithm, $$\begin{array}{rl}\text{gcd}(9n-2,7n+3)& =\text{gcd}(9n-2-(7n+3),7n+3)\\ & =\text{gcd}(2n-5,7n+3)\\ & =\text{gcd}(2n-5,7n+3-3(2n-5))\\ & =\text{gcd}(2n-5,n+18)\\ & =\text{gcd}(2n-5-2(n+18),n+18)\\ & =\text{gcd}(-41,n+18).\end{array}$$Since $41$ is prime, it follows that $9n-2$ and $7n+3$ have a common factor greater than 1 only if $n+18$ is divisible by 41. The smallest such positive integer value of $n$ is $41-18=\boxed{23}$. Note that $9n-2=205=5\times 41$ and $7n+3=164=4\times 41$.