Suppose we have a graph on $n$ nodes. We would like to assign to each node either a $+1$ or a $-1$. Call this a configuration $\sigma \in \{+1,-1\}^n$. The number of $+1$s that we have to assign is exactly $s$ (hence the number of $-1$s is $n-s$.) Given a configuration $\sigma$, we look at each node $i$ and sum the values assigned to its neighbors, call this $\xi\_i(\sigma)$. We then count the number of nodes for which $\xi\_i(\sigma)$ is nonnegative:


$$ N(\sigma) := \sum\_{i=1}^n 1( \xi\_i(\sigma) \ge 0). $$


The question is: what is the configuration $\sigma$ that maximizes $N(\sigma)$? Can we give a bound on $(\max N)/n$ in terms of $s/n$. If it helps, the graph can be assumed to be Erdős-Renyi. 



An optimization involving (random) graphs
OK. Let me make a wild guess. I'll assume that you're looking at a random graph with $p=(\log n)/n$ and $s=n/4$ for concreteness.


Since for $p$ in that range, there are very few small cycles, the graph is behaving quite a lot like a tree with branching number $\log n$ at each vertex. 


An easy lower bound for the number of vertices that you can "infect" is $2\gamma n/\log n$: pick out a set $S$ with this number of vertices arbitrarily; and then infect half of their neighbours. In fact you can clearly do better since some vertices are neighbours with several elements of $S$ - this means you can do extra infection for free.


My guess: $\max N\sim Cn/\log n$ (possibly up to something like $\log\log n$ factors). 


