Solution:

Let $N = 2011^{2011}$. Since $N > 17489$, there are more than $2011$ primes in the sequence $1, 2, 3, \ldots, N$.

Claim. There exists a sequence of $N$ consecutive positive integers that are all composite.

Proof of the Claim. The sequence
$$(N+1)! + 2,\ (N+1)! + 3,\ (N+1)! + 4,\ \ldots,\ (N+1)! + N,\ (N+1)! + (N+1)$$
consists of $N$ consecutive composite numbers, and so the claim follows.

Back to the solution of the problem, let
$$a,\ a+1,\ a+2,\ \ldots,\ a+N-1$$
be a sequence of $N$ consecutive positive integers with no prime. We do repeatedly the following operation to the numbers in $(\star)$. Delete the far-right number $a+N-1$, and append to the far-left the number $a-1$. The resulting sequence
$$a-1,\ a,\ a+1,\ \ldots,\ a+N-2$$
has at most one prime. Repeating this operation until we reach the sequence $1, 2, 3, \ldots, N$, which has more than $2011$ primes.

Performing such operation either retains, increases by one, or decreases by one the number of primes of the previous sequence. Since the starting sequence has no prime at all, while the last sequence has more than $2011$ primes, there exists a sequence (after applying the operation a number of times) that contains exactly $2011$ primes.