NP-hardness of finding 0-1 vector to maximize rows of {-1, +1} matrix
This problem is NP-hard. In fact, even deciding whether there exists a choice of x such that all the dot products are positive is NP-complete. I show this below by reduction from set cover.

Reduction

Suppose we are given a set cover instance consisting of number $k$ and sets $S_1, S_2, \ldots, S_m \subseteq \{1,2,\ldots,n\}$. The corresponding question is whether there exists a subset $C$ of $\{1,2,\ldots,n\}$ with $|C| = k$ such that each $S_i$ contains at least one element of $C$ (or equivalently such that $S_i \cap C$ is non-empty for each $i$).

We use this input to construct an instance of your problem, that is, a set of vectors over $\{-1, 1\}$ of the same length. The length of these vectors is going to be $n+k+1$, and there will be $m+k+2$ of them. In particular, we will have vectors $a$, $b_1, b_2, \ldots, b_{k+1}$, and $c_1, c_2, \ldots, c_m$ as defined below. 

Denote by $1(n)$ a vector of length $n$ whose values are all $1$s. Denote by $e_i(n)$ a vector of length $n$ whose $i$th value is $1$ and whose other values are $-1$. Denote by $s_j$ a vector of length $n$ whose $i$th value is $1$ if $i \in S_j$ and $-1$ otherwise.

Then


$a = [~1(k+1)~;~-1(n)~]$
$b_j = [~e_j(k+1)~;~1(n)~]$ for $1 \le j \le k+1$ and
$c_j = [~-e_1(k+1)~;~s_j~]$ for $1 \le j \le n$


Correctness

Let $x = [~x'~;~x''~]$ be any vector of $0$s and $1$s expressed as two subvectors $x'$ and $x''$ of lengths $k+1$ and $n$. 

Suppose that the dot product of $x$ with vectors $a$, $b_1$, $\ldots$, and $b_{k+1}$ is positive. Then consider the two vectors $a$ and $b_j$ (for some $j$). These two vectors are almost exactly negatives of each other. In particular, they both have $1$ as the $j$th component but are otherwise negatives of each other (this is easy to verify). Then if $x$ is a vector of $0$s and $1$s with a $0$ in the $j$th component, the dot product of $x$ with $a$ is the negative of the dot product of $x$ with $b_j$. In this case, these two dot products cannot both be positive. Therefore, $x$ must have a $1$ in the $j$th component.

The above logic can be applied for every $j$ with ($1 \le j \le k+1$). From this, we can conclude that if $x$ has a positive dot product with vectors $a$, $b_1$, $\ldots$, and $b_{k+1}$, then $x' = 1(k+1)$.

Furthermore, we know that $x \cdot a &gt; 0$, and $$x \cdot a = [~x'~;~x''~] \cdot [~1(k+1)~;~-1(n)~]= [~1(k+1)~;~x''~] \cdot [~1(k+1)~;~-1(n)~] = 1(k+1) \cdot 1(k+1) + x'' \cdot -1(n) = k+1 - |x''|_1$$ where $|x''|_1$ is the number of $1$s in $x''$, so $k+1-|x''|_1 &gt; 0$ and $|x''|_1 &lt; k+1$.

Similarly, $x \cdot b_1 &gt; 0$, and $$x \cdot b_1 = [~x'~;~x''~] \cdot [~e_1(k+1)~;~1(n)~]= [~1(k+1)~;~x''~] \cdot [~e_1(k+1)~;~1(n)~] = 1(k+1) \cdot e_1(k+1) + x'' \cdot 1(n) = -(k-1) + |x''|_1,$$ so $-(k-1)+|x''|_1 &gt; 0$ and $|x''|_1 &gt; k-1$.

Together the above implies that $|x''|_1 = k$. To summarize, we have shown that if $x$ has a positive dot product with vectors $a$, $b_1$, $\ldots$, and $b_{k+1}$, then $x' = 1(k+1)$ and $|x''|_1 = k$. In fact, it is easy to verify that the reverse is also true. In other words, $x = [~x'~;~x''~]$ has a positive dot product with vectors $a$, $b_1$, $\ldots$, and $b_{k+1}$ if and only if $x' = 1(k+1)$ and $|x''|_1 = k$. 

From now on, consider only vectors $x$ of the above form. Notice that there is a one to one correspondence between possible vectors $x''$ and subsets of $\{1, \ldots, n\}$ of size $k$: in particular, $x''$ can be put into correspondence with set $X \subseteq \{1, \ldots, n\}$ with $|X| = k$ such that $(x'')_i = 1$ iff $i \in X$.

Under this correspondence, we can consider the necessary and sufficient conditions for $x$ to have positive dot product with vectors $c_1$, $\ldots$, and, $c_n$.

For any $j$, we can expand $x \cdot c_j$ as $$x \cdot c_j = [~1(k+1)~;~x''~] \cdot [~-e_1(k+1)~;~s_j~] = 1(k+1) \cdot -e_1(k+1) + x'' \cdot s_j = (k-1) + x'' \cdot s_j.$$
Notice that the value of $x'' \cdot s_j$ is exactly a sum of $k$ components of $s_j$: in particular, $x'' \cdot s_j = \sum_{i \in X}(s_j)_i$. If these $k$ components are all $-1$s, then this sum is $-k$, and so $x \cdot c_j = (k-1) + x'' \cdot s_j = (k-1) + -k = -1 &lt; 0$. If these $k$ components include at least one $1$, then this sum is at least $-1\times(k-1) + 1\times 1 = -(k-2)$ in which case $x \cdot c_j = (k-1) + x'' \cdot s_j \ge (k-1) + -(k-2) = 1 &gt; 0$. Thus, we see that $x \cdot c_j \ge 0$ for a specific $j$ if and only if $X$ includes at least one $i$ such that $(s_j)_i = 1$. 

Remember that $(s_j)_i = 1$ if and only if $i \in S_j$. Thus, $x \cdot c_j \ge 0$ for a specific $j$ if and only if $X$ includes at least one $i$ such that $i \in S_j$, or equivalently if and only if $X \cap S_j$ is non-empty. 

Then $x \cdot c_j \ge 0$ for every $j$ if and only if for every $j$, $X \cap S_j$ is not empty. Notice that the latter condition is exactly equivalent to the condition that $X$ is a solution to the input set cover instance.

In summary, we see that $x$ is a vector of $0$s and $1$s with positive dot product with all the vectors produced by the reduction if and only if $x = [~1(k+1)~;~x''~]$ with $|x''|_1 = k$ and $x''$ corresponds to a set $X \subseteq \{1, \ldots, n\}$ with $|X| = k$ such that $X$ is a solution to the input instance of set cover. In other words, a solution to the set cover instance exists if and only if a solution to the instance produced by the reduction exists.