Must be a popular lecturer

Since there are $19$ girls seated in each row there are $19r$ chairs occupied by girls, and since there are $15$ boys seated in each column there are $15c$ chairs occupied by boys. There are a total of $rc$ chairs in the lecture hall, so with $14$ empty chairs we can form the equation

$19r + 15c + 14 = rc \Longrightarrow 15c + 14 = r(c - 19)$

$\Longrightarrow r = \dfrac{15c + 14}{c - 19} = \dfrac{15c - (15)(19) + (15)(19) + 14}{c - 19} = 15 + \dfrac{299}{c - 19}.$

Now $r$ must be an integer so $c - 19$ must divide $299.$ Since $299 = 13*23,$ we can have $c - 19$ equal $1, 13, 23$ or $299.$ This gives us four possible pairs $(c,r)$ , namely

$(20, 314), (32,38), (42,28), (318, 16).$

The resulting products are $6280, 1216, 1176, 5088,$ respectively, and so the desired minimum of $r \times c$ is $\boxed{1176}.$