For every $4$ seats you need to keep at least $1$ open to not have $4$ consecutive chairs occupied. So divide the row in sets $S_k = \{4k + 1, 4k + 2, 4k + 3, 4k + 4\}$ for $k = 0, \ldots, 7$ and $S_8 = \{33, 34, 35\}$. For each set $S_0, \ldots, S_7$ you need to keep one seat open, so you need at least $8$ open seats to not have a sequence of $4$ occupied seats. This maximum can also be achieved, by leaving seats open at positions $4k$, for $k = 1, \ldots, 8$.
With respect to applying the pigeonhole principle: If you do have more than $35 - 8 = 27$ seats filled, then you have at least $25$ seats filled for $S_0, \ldots, S_7$. Since $25 / 8 > 3$, by the pigeonhole principle one of them must have at least $4$ seats occupied. But then you get a sequence of $4$ occupied seats. So if $28$ or more seats are occupied, you always have $4$ or more consecutive occupied seats.