We claim that the answer is when $3 \nmid n$ and $n \neq 2.$ If $3|n$, then if we let $a_i = 1$ if the $i$th lightbulb is on and $a_i = 0$ else, then $a_1 a_2 + a_4-a_5 + \cdots + a_{n-2} - a_{n-1}$ is invariant modulo $2$. Therefore, if we simply let the $1$st lightbulb be on and everything else be off initially, there's no way to get everything off. If $n = 2$, then let one be one, the other be off initially, there's clearly no way to make everything off. Hence, it only suffices to show that when $n>2$ and $3 \nmid n$, we can always turn all lights off. Let's interpret the problem in $F_2$. Define the vectors $v_1 = (1, 1, 0, \cdots, 1), v_2 = (1, 1, 1, 0, 0, \cdots, 0), v_3 = (0, 1, 1, 1, 0, \cdots, 0), \cdots v_n = (1, 0, \cdots, 0, 1, 1).$ If the vector corresponding to our state is currently $v$, then we can add $v_i$ if the $i$th coordinate of $v$ is $1$. Our goal is to eventually make $v$ the zero vector. Lemma. The $v_i$'s form a basis of $\{0, 1\}^n$ in $F_2$. Proof. Just show that there is no nonzero kernel. This is very easy to do. $\blacksquare$ With the lemma, we can associate every vector $v \in \{0, 1\}^n$ with a "todo-list," which is a set of indices from $1$ to $n$, so that if we add $v_i$ for those $i$'s to this vector $v$, then it becomes the zero vector. It's clear that every vector $v$ has a unique todo-list $T \subseteq \{1, 2, \cdots, n\}$ and that $v = \sum_{i \in T} v_i$. We will describe an algorithm which strictly decreases the todo-list of a given state vector $v$ using only permissible moves. First of all, if there exists an $i \in T$ so that the $i$th coordinate of $v$ is $1$, then we simply add $v_i$ to $v$ (this is allowed), which changes the todo-list from $T$ to $T / \{i\}.$ Otherwise, suppose that for all $i \in T$, we have that the $i$th coordinate of $v$ is zero. This implies that for $i \in T$, exactly one of $\{i-1, i+1\}$ is also in $T$. Therefore, we have that the indices in $T$ "come in adjacent pairs," in the sense that we can pair up the elements of $T$ into adjacent pairs of indices which are non-touching. Call these adjacent pairs blocks . Since $3 \nmid n$, there must exist two neighboring blocks for which the arc between them contains more than one index. Say that the block $\{i, i+1\}$ and the block $\{j, j+1\}$ satisfy that $i+2, i+3, \cdots, j-1 \notin T$ and $j \equiv i+k$ mod $n$ where $3 < k < n.$ Then, notice that the $(i+2)$th coordinate of $v$ is $1$, and so we can add $v_{i+2}$, then $v_{i+1}$ (since adding $v_{i+2}$ would've changed the $(i+1)$th coordinate to $1$), and then $v_{i+2}$ again. This removes $i+1$ from our todo-list, hence decreasing the size of the todo-list. As we've described an algorithm which eventually makes the todo-list the empty set starting from any state vector $v$, we're done. $\square$