Expected number of numbers to be picked from $[0, 2n - 1]$ with repetition until two numbers sum to $2n$. Given we are picking numbers from $[0, 2n - 1]$ with repetition, how many numbers should we need to pick on the average, such that some two picked numbers sum up to $2n$. 
Here is a simplified version of the question. Given we are picking numbers from the set $S = \{0, 1, 2, 3\}$ with repetition. How many numbers should we pick on the average, such that some two picked numbers sum up to 4. Monte-Carlo simulation shows that the answer is 4.222. I have no idea how to compute this analytically.
(Case n = 1) Consider the case when we are pick numbers with repetition from $S = \{0, 1\}$. For the expected number of numbers to be picked until some two picked numbers sum up to 2. We can compute the expectation like this:
$E = \frac{1}{2}(1 + E) + \frac{1}{2}(\frac{1}{2}(2) + \frac{1}{2}(1 + E_1))$, where E is the desired expectation and $E_1$ is the expected wait time until you draw a 1. So $E_1 = 2$, and $E = 4$.

Let $E$ be the expected number of numbers drawn from $[0, 2n - 1]$ until two of the picked numbers
sum up to $2n$. Let $E_{x,y}$ be the expected number of numbers to be drawn until two of the
picked numbers sum up to $2n$, where $x$ is the number of distinct numbers seen so far belonging
to the set $S = \{1, 2, 3, ..., (n - 1), (n + 1), ..., (2n - 1)\}$. And, $y \in \{1, 2, 3, 4\}$ where 1 - no $0$ is seen, no $n$ is seen, 2 - $0$ is seen, no $n$ is seen, 3 - no $0$ is seen, $n$ is seen, 4 - $0$ is seen, $n$ is seen. 
Then,
$$E = E_{0,1}$$
where,
$$E_{i,1} = \frac{1}{2n}(1 + E_{i,2}) + \frac{1}{2n}(1 + E_{i,3}) + \frac{i}{2n}(1 + E_{i,1}) + \frac{i}{2n}(1) + \frac{2n - 2i - 2}{2n}(1 + E_{i+1,1})$$
$$E_{i,2} = \frac{1}{2n}(1 + E_{i,2}) + \frac{1}{2n}(1 + E_{i,4}) + \frac{i}{2n}(1 + E_{i,2}) + \frac{i}{2n}(1) + \frac{2n - 2i - 2}{2n}(1 + E_{i+1,2})$$
$$E_{i,3} = \frac{1}{2n}(1 + E_{i,4}) + \frac{1}{2n}(1) + \frac{i}{2n}(1 + E_{i,3}) + \frac{i}{2n}(1) + \frac{2n - 2i - 2}{2n}(1 + E_{i+1,3})$$
$$E_{i,4} = \frac{1}{2n}(1 + E_{i,4}) + \frac{1}{2n}(1) + \frac{i}{2n}(1 + E_{i,4}) + \frac{i}{2n}(1) + \frac{2n - 2i - 2}{2n}(1 + E_{i+1,4})$$
The base cases are $E_{n,1} = 0, E_{n,2} = E_{n,3} = 0, E_{n,4} = 0$.
To solve this problem, all we need to do is to setup these equations and solve them. The equations can be solved simply by back-substitution. I have manually verified the answers for $n = 1$, $n = 2$ and $n = 3$.
