Solution:

Answer. $k=34$.

Preliminary notes. Let $F$ denote the number of occupied chairs at the current position in the game. Notice that, on any turn, $F$ does not decrease. Thus, we need to determine the maximal value of $F$ Ann can guarantee after an arbitrary move (either hers or her opponent's).

Say that the situation in the game is stable if every unoccupied chair is adjacent to an occupied one. In a stable situation, we have $F \geq 33$, since at most $3F$ chairs are either occupied or adjacent to such. Moreover, the same argument shows that there is a unique (up to rotation) stable situation with $F=33$, in which exactly every third chair is occupied; call such stable situation bad.

If the situation after Bob's move is stable, then Bob can act so as to preserve the current value of $F$ indefinitely. Namely, if $A$ puts some guest on chair $a$, she must free some chair $b$ adjacent to $a$. Then Bob merely puts a guest on $b$ and frees $a$, returning to the same stable position.

On the other hand, if the situation after Bob's move is unstable, then Ann may increase $F$ in her turn by putting a guest on a chair having no adjacent occupied chairs.

Strategy for Ann, if $k \leq 34$. In short, Ann's strategy is to increase $F$ avoiding appearance of a bad situation after Bob's move (conversely, Ann creates a bad situation in her turn, if she can).

So, on each her turn, Ann takes an arbitrary turn increasing $F$ if there is no danger that Bob reaches a bad situation in the next turn (thus, Ann always avoids forcing any guest to stand up). The exceptional cases are listed below.

Case 1. After possible Ann's move (consisting in putting a guest on chair $a$), we have $F=32$, and Bob can reach a bad situation by putting a guest on some chair. This means that, after Ann's move, every third chair would be occupied, with one exception. But this means that, by her move, Ann could put a guest on a chair adjacent to $a$, avoiding the danger.

Case 2. After possible Ann's move (by putting a guest on chair $a$), we have $F=33$, and Bob can reach a stable situation by putting a guest on some chair $b$ and freeing an adjacent chair $c$. If $a=c$, then Ann could put her guest on $b$ to create a stable situation after her turn; that enforces Bob to break stability in his turn. Otherwise, as in the previous case, Ann could put a guest on some chair adjacent to $a$, still increasing the value of $F$, but with no danger of bad situation arising.

So, acting as described, Ann increases the value of $F$ on each turn of hers whenever $F \leq 33$. Thus, she reaches $F=34$ after some her turn.

Strategy for Bob, if $k \geq 35$. Split all chairs into 33 groups each consisting of three consecutive chairs, and number the groups by $1,2, \ldots, 33$ so that Ann's first turn uses a chair from group 1. In short, Bob's strategy is to ensure, after each his turn, that

$(*)$ In group 1, at most two chairs are occupied; in every other group, only the central chair may be occupied.

If $(*)$ is satisfied after Bob's turn, then $F \leq 34 < k$; thus, property $(*)$ ensures that Bob will not lose. It remains to show that Bob can always preserve $(*)$ after any his turn. Clearly, he can do that at the first turn.

Suppose first that Ann, in her turn, puts a guest on chair $a$ and frees an adjacent chair $b$, then Bob may revert her turn by putting a guest on chair $b$ and freeing chair $a$.

Suppose now that Ann just puts a guest on some chair $a$, and the chairs adjacent to $a$ are unoccupied. In particular, group 1 still contains at most two occupied chairs. If the obtained situation satisfies $(*)$, then Bob just makes a turn by putting a guest into group 1 (preferably, on its central chair), and, possibly, removing another guest from that group. Otherwise, $a$ is a non-central chair in some group $i \geq 2$; in this case Bob puts a guest to the central chair in group $i$ and frees chair $a$.

So Bob indeed can always preserve $(*)$.