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Consider an nitrogen atom with 7 protons and 7 electrons. How can an nitrogen anion $\ce{N^-}$ exist? Shouldn't the 7 electrons in valence shell repel the extra one?
What force does hold the extra one electron in the valence shell? There isn't an extra proton in the $\ce{N^-}$ that would hold the extra electron.
One reason cations and anions exist is due to the stability of a full or half-full valence shell. The stability from those electronic configurations means that the atom or molecule does not require protons to "hold" the extra electron.
Recall also that nitrogen has three (or five) valence electrons, rather than seven. The 1s shell is full and is not considered part of its valency. The three 2p electrons are the valence electrons although they hybridize with the 2s electrons to produce the trigonal pyramidal structure of ammonia with its lone pair.
The single anion $\ce{N^-}$ could exist, but would not be stable because it puts four electrons in the p shell. The p shell would prefer to have three electrons as it does in the nitrogen atom or no electrons as it does in the $\ce{N^3+}$ cation.
Something to consider: does N become an anion all by itself? No. It gets those electrons from a much less electronegative atom. When it does that it generates an anion and a cation. They attract each other. Potential energy decreases (energy is released) and that "released" energy drives the REACTION. As I like to say to students: It's not like a sodium atom is walking down the street and one of its electrons falls out of its pocket. Metals don't just LOSE electrons. That electron has to be TAKEN by a more electronegative atom.
