Suppose a list $L = (n_i)_{i=1}^N$ is made up only of numbers selected from the (possibly much shorter) list of distinct numbers $(m_i)_{i=1}^k$, and that the products $n_i n_{i+1}$, $1 \le i \le 2010$, are all distinct. We write down the indices of the numbers $m_i$ occurring in $L$ as another list $M$ also of length $N$, e.g. if $L$ begins $m_3, m_7, m_1, m_9, m_7, \dots$, then $M$ begins $3, 7, 1, 9, 7, \dots$.
If $n_i = m_a$ and $n_{i+1} = m_b$, the product $n_i n_{i+1}$ is determined by the set $\{a, b\}$. Hence, any two (not necessarily distinct) numbers $a, b$ from $\{1, 2, \dots, k\}$ can appear as neighbours in the list $M$ at most once. As $M$ contains $2011$ elements, there must exist at least $2010$ different subsets of $\{1, 2, \dots, k\}$ with one or two elements. Note that, by usual conventions of set theory, $\{a, a\} = \{a\}$ contains only one element. The number of such subsets is equal to $k(k+1)/2$ and so we need to have $k(k+1)/2 \ge 2010$, i.e. $k \ge 63$.
That $k=63$ is indeed sufficient follows from the construction of a sequence $M$ given below. We will in fact construct for each odd number $k \ge 3$ a 'cycle' of numbers such that each subset of $\{1, 2, \dots, k\}$ with one or two elements appears exactly once as a set of neighbours in the cycle. A 'cycle' of numbers is simply a list in which the first and last element coincide. Two such 'cycles' of length $T+1$ are said to be equivalent iff we obtain the same result, up to rotation, if their numbers are cyclically written at the vertices of a regular $T$-gon, whereby the first and last element of the list are written on the same vertex.
To prove the existence of such a cycle, we first form the set $\mathcal{T}$ of all subsets of $\{1, 2, \dots, k\}$ with one or two elements. Note that each number appears $k-1$ times in a two-element subset and once in a singleton. We start the sequence with $1, 2, 3, \dots, k$ and remove from $\mathcal{T}$ the sets $\{1, 2\}, \{2, 3\}, \dots, \{k-1, k\}$ which appear as neighbours already. We repeatedly do now the following:
* If $a$ is the last element of our list and there is a set in $\mathcal{T}$ which contains $a$, we pick such a set, remove it from $\mathcal{T}$ and append $a$ (if the set was a singleton) or the second element of this set (if it was a two-element set) to the list.
* Because $k-1$ is even, if $\mathcal{T}$ is not empty but does not contain a set which contains the last element of our list, then the last element of our list must coincide with the first element of our list, i.e. the list is a cycle. In this case, we replace it by an equivalent cycle which has an element as its first and last element which appears in one of the sets in $\mathcal{T}$. This is possible because we started our list in such a way that it contains all numbers $1, \dots, k$ at least once. After replacing the cycle by an appropriate equivalent one, we continue as above appending elements to the list.
This process can be continued until all elements of $\mathcal{T}$ are used. The result is a list which contains all possible neighbour-subsets exactly once. If $k=63$, the list will have length $63 \cdot 64/2 + 1 = 2017$. By discarding the last $6$ elements we obtain a list $M$ with $2011$ elements.
Let $M = (\mu_i)_{i=1}^{2011}$ be this list and let $m_i$ be the $i$-th prime number. Defining $n_i = m_{\mu_i}$ we obtain a list $L = (n_i)_{i=1}^{2011}$ which satisfies the conditions of the problem.