Compute the sum of all $n$ for which the equation $2x+3y=n$ has exactly $2011$ non-negative $(x,y ≥ 0)$ integer solutions. 
QUESTION: Compute the sum of all $n$ for which the equation $2x + 3y = n$ has exactly $2011$ non-negative $(x,y ≥ 0)$ integer solutions.


MY APPROACH: I am genuinely stumped by this one.. We are given an equation of a line.. A set of parallel lines to be precise.. But from where do I even start to solve something with so less information?
I am not asking to spoon-feed me but some hints will be really helpful here.. Is there some theory upon which this question is based?
Also I couldn't think of some relevant tags for this question.. If anyone wishes to add or alter tags is most welcome to do so..
Thanks in advance for your kind help.. :)
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A: Hint
Perhaps this can help you start.
Since the $gcd(2,3)=1$, so for every $n$ we will have integer solutions for the equation $2x+3y=n$ because.
$$2(2)+3(-1)=1 \implies 2(2n)+3(-n)=n.$$
But we are looking for $x,y \in \Bbb{Z}_{\geq 0}$. So we can go back to what we have above and observe that
$$2(2n)+3(-n)=n \implies 2\underbrace{(2n-3t)}_{\color{red}{x}}+3\underbrace{(-n+2t)}_{\color{red}{y}}=n \qquad \forall t \in \Bbb{Z}.$$
So we want $t$ such that
\begin{align*}
2n-3t & \geq 0\\
-n+2t & \geq 0.
\end{align*}
This gives us $$\frac{n}{2} \leq t \leq \frac{2n}{3}.$$
Now you want the number of such $t$'s to be exactly $2011$. See if this can give you clues about what kind of $n$ will do that.
Added explanation (in response to the comment below by @Stranger Forever)
We have to count the number of integers $t$ in the interval $\left[\frac{n}{2}, \frac{2n}{3}\right]$. The count will depend on the values that $n$ will take. It will not be simply $\frac{n}{6}$ as you have stated in your comment. In fact,
$$\# \text{ of integers in the given interval }=\begin{cases}\left\lfloor \frac{n}{6} \right\rfloor +1 & \text{ if } n \not\equiv 1 \pmod{6}\\ \left\lfloor \frac{n}{6} \right\rfloor  & \text{ if } n \equiv 1 \pmod{6}\end{cases}. $$
Since we want the number of solutions $(x,y)$ to be exactly $2011$, this means we want this count to be $2011$. Recall that $\lfloor a \rfloor =k \implies k \leq a < k+1$. Consequently, we get
$$\begin{cases}2010 \leq \frac{n}{6}  < 2011 & \text{ with } n \not \equiv 1 \pmod{6}\\ 2011 \leq \frac{n}{6}  < 2012 & \text{ with } n \equiv 1 \pmod{6}.\end{cases}$$
The first inequality gives us: $n \in \{12060, 1206\color{green}{2}, 12063,12064,12065\}$ (note the missing $\color{red}{12061}$ as it is $1 \bmod{6}$).
The second inequality gives us: $n=12067$ (as it will be the ONLY number that is $1 \bmod 6$ in that interval.)
The problem wants the sum of these possible values of $n$. So we want
$$12060+12062+12063+12064+12065+12067=\boxed{\color{red}{72381}}.$$
