Let $a$ and $b$ be the two possible values of $\tan\theta$ given that \[\sin\theta + \cos\theta = \dfrac{193}{137}.\] If $a+b=m/n$, where $m$ and $n$ are relatively prime positive integers, compute $m+n$.
1. Given the equation \(\sin\theta + \cos\theta = \dfrac{193}{137}\), we start by squaring both sides to utilize the Pythagorean identity.
   \[
   (\sin\theta + \cos\theta)^2 = \left(\dfrac{193}{137}\right)^2
   \]
   \[
   \sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta = \dfrac{193^2}{137^2}
   \]
   Using the Pythagorean identity \(\sin^2\theta + \cos^2\theta = 1\), we get:
   \[
   1 + 2\sin\theta\cos\theta = \dfrac{193^2}{137^2}
   \]
   \[
   2\sin\theta\cos\theta = \dfrac{193^2}{137^2} - 1
   \]
   \[
   2\sin\theta\cos\theta = \dfrac{193^2 - 137^2}{137^2}
   \]
   \[
   2\sin\theta\cos\theta = \dfrac{(193 - 137)(193 + 137)}{137^2}
   \]
   \[
   2\sin\theta\cos\theta = \dfrac{56 \cdot 330}{137^2}
   \]
   \[
   2\sin\theta\cos\theta = \dfrac{18480}{18769}
   \]
   \[
   \sin\theta\cos\theta = \dfrac{9240}{18769}
   \]

2. Recall that \(\sin\theta\cos\theta = \dfrac{1}{2}\sin(2\theta)\), so:
   \[
   \dfrac{1}{2}\sin(2\theta) = \dfrac{9240}{18769}
   \]
   \[
   \sin(2\theta) = \dfrac{18480}{18769}
   \]

3. We need to find \(\tan\theta\). Using the identity \(\tan(2\theta) = \dfrac{2\tan\theta}{1 - \tan^2\theta}\), let \(\tan\theta = t\). Then:
   \[
   \sin(2\theta) = \dfrac{2t}{1 + t^2}
   \]
   \[
   \dfrac{2t}{1 + t^2} = \dfrac{18480}{18769}
   \]
   Cross-multiplying gives:
   \[
   2t \cdot 18769 = 18480(1 + t^2)
   \]
   \[
   37538t = 18480 + 18480t^2
   \]
   Rearranging terms:
   \[
   18480t^2 - 37538t + 18480 = 0
   \]

4. Solving this quadratic equation using the quadratic formula \(t = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 18480\), \(b = -37538\), and \(c = 18480\):
   \[
   t = \dfrac{37538 \pm \sqrt{37538^2 - 4 \cdot 18480 \cdot 18480}}{2 \cdot 18480}
   \]
   \[
   t = \dfrac{37538 \pm \sqrt{1409157444 - 1367424000}}{36960}
   \]
   \[
   t = \dfrac{37538 \pm \sqrt{41733444}}{36960}
   \]
   \[
   t = \dfrac{37538 \pm 6459}{36960}
   \]
   This gives us two solutions:
   \[
   t_1 = \dfrac{37538 + 6459}{36960} = \dfrac{43997}{36960}
   \]
   \[
   t_2 = \dfrac{37538 - 6459}{36960} = \dfrac{31079}{36960}
   \]

5. Adding the two solutions:
   \[
   a + b = \dfrac{43997}{36960} + \dfrac{31079}{36960} = \dfrac{75076}{36960}
   \]
   Simplifying the fraction:
   \[
   \dfrac{75076}{36960} = \dfrac{18769}{9240}
   \]

6. Since \(a + b = \dfrac{18769}{9240}\), and \(m\) and \(n\) are relatively prime, we have \(m = 18769\) and \(n = 9240\). Therefore, \(m + n = 18769 + 9240 = 28009\).

The final answer is \(\boxed{28009}\)