When states that are not accepting states become accepting states in NFA, what happens?
It depends on whether the NFA contains a path for every possible word.

If there is an input word where there is no path through the NFA at all labelled with that word, then yes, this can happen.  See Hendrik Jan's answer.

On the other hand, it is often assumed that the automaton is complete: that for every possible input word, there is at least one path in the automaton labelled with that word.  (For instance, this can be arranged by ensuring that for every state $q$ and every possible input letter $x \in \Sigma$, there is a transition out of $q$ labelled with $x$.)  If you assume the automaton is complete, the answer is: No.

In particular, if the automaton is complete, the answer is: No, the magic of nondeterminism is that if there's any valid computation path that will end in an accept state, the string is accepted. In this case, all possible computation paths in the original NFA led to rejection (had there been any valid path the string would be accepted, so they must all reject), so in the inverted NFA, they all lead to acceptance.