Answer. Yes.

Imagine a $32 \times 5$ table $T$ such that the set of its rows is equal to the set of all 5-tuples $(b_0, b_1, \ldots, b_4)$ in which $\forall\ 0 \le i \le 4 : b_i \in \{0,1\}$.

Now consider a $32 \times 16$ table $R$. Let $(c_i)$ be the $i$-th column and the $j$-th column $(c_j)$ is "consecutive" if and only if $i - j \equiv \pm 1 \pmod{16}$.

Show the square located in the row $r_i$ and the column $c_j$ with the sign $(i, j)$. Also all the tables made of the first, second and third five columns of $R$ as $T_1, T_2$ and $T_3$ respectively.

Then we will fill the table using the following method: Put a copy of table $T$ in the squares of table $T_1$ (which means that the values of the corresponding squares in $T$ and $T_1$ are the same.) And do the same for table $T_2$ and $T_3$. Then the only empty squares of $R$ are the squares in $c_{16}$.

If we put $\omega_{i,16} \equiv \sum_{j=1}^{5} \omega_{i,j} \pmod{2}$ (1), in which $\omega_{i,j}$ means the value of $(i, j)$, then there will be an answer for the problem if we stick $R$ around a cylinder.

Proof. Assuming $S$ as a set of 5 "consecutive" columns in $R$, there can be two conditions:

i) $c_{16} \notin S$: Then we knew that $S$ contains 5 different columns in $S$ is "nice".

ii) $c_{16} \in S$: The other 4 columns in $S$ (except for $c_{16}$,) are copies of 4 different columns in $T$, i.e. assume that they are similar to $c_1, c_2, c_3$ and $c_4$ (2). If there exist two rows $r_i$ and $r_j$ in $S$ ($i \neq j$) with equal values, then using (2) we will have: $\forall\ 1 \le k \le 4: \omega_{i,k} = \omega_{j,k} \Rightarrow$ According to the property that $T$ is nice, $\omega_{i,5} \neq \omega_{j,5} \Rightarrow s_i \neq s_j$, in which $s_i$ is the sum of $i$-th row in $T$. But $r_i$ and $r_j$ are similar so $\omega_{i,16} = \omega_{j,16} \stackrel{(1)}{\implies} s_i = s_j$ which is a contradiction. ■