Solution:

Since $AK = AL = KL = 14,$ triangle $AKL$ is equilateral and $\angle KAL = 60^\circ.$ Let $\alpha = \angle BAK$ and $\beta = \angle LAC,$ so $\alpha + \beta = 30^\circ.$ Because $AK = KB,$ point $K$ lies on the perpendicular bisector of $\overline{AB},$ so $AB = 2 \cdot 14\cos\alpha = 28\cos\alpha;$ similarly $AC = 28\cos\beta.$ Then $AB^2 + AC^2 = 38^2$ gives $\cos^2\alpha + \cos^2\beta = \frac{361}{196},$ i.e. $\cos 2\alpha + \cos 2\beta = \frac{165}{98}.$ By sum-to-product, $2\cos(\alpha+\beta)\cos(\alpha-\beta) = \sqrt{3}\cos(\alpha - \beta) = \frac{165}{98},$ so $\cos(\alpha - \beta) = \frac{55\sqrt{3}}{98}.$

Decompose $[BKLC] = [ABC] - [ABK] - [ACL] - [AKL].$ First, $$[ABC] = \tfrac{1}{2} AB \cdot AC = 392\cos\alpha\cos\beta = 196\bigl(\cos(\alpha - \beta) + \cos(\alpha + \beta)\bigr) = 110\sqrt{3} + 98\sqrt{3} = 208\sqrt{3}.$$ Next, $K$ has height $14\sin\alpha$ over $\overline{AB},$ so $[ABK] = \frac{1}{2} \cdot 28\cos\alpha \cdot 14\sin\alpha = 98\sin 2\alpha,$ and likewise $[ACL] = 98\sin 2\beta;$ their sum is $196\sin(\alpha + \beta)\cos(\alpha - \beta) = 98 \cdot \frac{55\sqrt{3}}{98} = 55\sqrt{3}.$ Finally $[AKL] = \frac{\sqrt{3}}{4} \cdot 14^2 = 49\sqrt{3}.$

Therefore $[BKLC] = 208\sqrt{3} - 55\sqrt{3} - 49\sqrt{3} = 104\sqrt{3},$ so $n = 104.$