Let $a$ and $b$ be nonzero real numbers such that $\tfrac{1}{3a}+\tfrac{1}{b}=2011$ and $\tfrac{1}{a}+\tfrac{1}{3b}=1$.  What is the quotient when $a+b$ is divided by $ab$?
1. Let $\dfrac{1}{a} = p$ and $\dfrac{1}{b} = q$. Then the given equations can be rewritten as:
   \[
   \dfrac{1}{3}p + q = 2011
   \]
   \[
   p + \dfrac{1}{3}q = 1
   \]

2. We are looking for the value of $\dfrac{a+b}{ab}$. Notice that:
   \[
   \dfrac{a+b}{ab} = \dfrac{1}{a} + \dfrac{1}{b} = p + q
   \]

3. To find $p + q$, we add the two equations:
   \[
   \dfrac{1}{3}p + q + p + \dfrac{1}{3}q = 2011 + 1
   \]
   Simplifying the left-hand side:
   \[
   \dfrac{1}{3}p + p + q + \dfrac{1}{3}q = 2012
   \]
   \[
   \left(\dfrac{1}{3} + 1\right)p + \left(1 + \dfrac{1}{3}\right)q = 2012
   \]
   \[
   \dfrac{4}{3}p + \dfrac{4}{3}q = 2012
   \]
   Factoring out $\dfrac{4}{3}$:
   \[
   \dfrac{4}{3}(p + q) = 2012
   \]

4. Solving for $p + q$:
   \[
   p + q = \dfrac{2012 \cdot 3}{4}
   \]
   \[
   p + q = \dfrac{6036}{4}
   \]
   \[
   p + q = 1509
   \]

Conclusion:
\[
p + q = \boxed{1509}
\]