Given that a positive integer is 1 less than another, we can assume the smaller integer to be x-1 and the larger integer to be x.
Let's consider the sum of the reciprocal of the smaller and twice the reciprocal of the larger:
\frac{1}{(x-1)} + 2\cdot\frac{1}{x} = \frac{19}{42}
Multiplying through by 42x(x-1) to clear the fractions:
42x + 84(x-1) = 19x(x-1)
Expanding and simplifying:
42x + 84x - 84 = 19x^2 - 19x
126x - 84 = 19x^2 - 19x
Rearranging terms:
19x^2 - 145x + 84 = 0
Now, we need to factor the quadratic equation or use the quadratic formula to solve for x.
After factoring or solving with the quadratic formula, we find that the two integers are 6 and 7
\boxed{x=7} and \boxed{x-1=6}