Solution:

For each value $v = b - a,$ the pairs $(a, b) = (1, v+1), (2, v+2), \ldots, (20-v, 20)$ show that $v$ occurs exactly $20 - v$ times, so $K = \prod_{v=1}^{19} v^{20-v}.$ The exponent of $2$ in $K$ is therefore $\sum_v (20 - v)\,e(v),$ where $e(v)$ is the exponent of $2$ in $v.$

Only even $v$ contribute: $v = 2, 6, 10, 14, 18$ give $e = 1;$ $v = 4, 12$ give $e = 2;$ $v = 8$ gives $e = 3;$ and $v = 16$ gives $e = 4.$ The total is $$18 + 16 \cdot 2 + 14 + 12 \cdot 3 + 10 + 8 \cdot 2 + 6 + 4 \cdot 4 + 2 = 150,$$ so $n = 150.$