NP-Complete problem proof
Consider an instance $G=(V,E)$ of vertex cover with $n=|V|$ and $m=|E|$, and build a directed graph $G' = (V', E')$ in which:


$V' = V \cup ( E \times \{0, 1, \dots, n\} )$.
For each $e = (u,v) \in E$, $E'$ contains all edges $(u,e')$ and $(v,e')$ for $e' \in \{e\} \times \{0, 1, \dots, n\}$.


Claim: If there is a set of at most $k$ observers that reaches at least $(n+1)m$ people in $G'$, then there is a vertex cover of size at most $k$ in $G$.

Proof:
The existence of a set a set of $k$ observers that reaches at least $(n+1)m$ people in $G'$
implies the existence of a set $S \subseteq V$ of at most $k$ observers that reaches at least $(n+1)m$ people in $G'$.

For every edge $e=(u,v) \in E$,  $\{u,v\} \cap S \neq \emptyset$. 
Indeed, if $\{u,v\} \cap S = \emptyset$, then none of the vertices in $e \times \{0, \dots, n\}$ is reached by $S$ in $G'$, showing that $S$ reaches at most $(n+1)m+n - (n+1) = (n+1)m-1$ people in $G'$.
Therefore $S$ is a vertex cover for $G$. $\square$

Claim: If there is a vertex cover $C$ of size at most $k$ in $G$, then there is a set of at most $k$ observers that reaches at least $(n+1)m$ people in $G'$.

Proof:
For each edge $e = (u,v) \in E$, let $x \in \{u,v\} \cap S$. All vertices $\{e\} \times \{0, \dots, n\}$ in $G'$ are reached by $x$ and hence by $C$. 
Therefore $C$ reaches at least $m \cdot (n+1)$ people in $G$. $\square$

This shows that your problem is NP-hard.