If I can't draw resonance structures for a pi system, does that mean it doesn't exist? Whilst trying to answer this question I came across an interesting situation regarding the relationship between resonance structures and molecular orbitals. There are two places that you can deprotonate this compound. The two different deprotonated forms are shown below along with their resonance structures. 

Both compounds have a continuous system of p orbitals which can align in a plane and overlap to form a $\pi$ system. However, in the first example it is not possible to draw resonance structures that delocalise the negative charge over all five atoms that have the available p orbitals. Does this mean that the negative charge is not delocalised over all five atoms? @Martin said in his comment on this answer that valence bond theory and molecular orbital theory ultimately give the same result so this should also apply here. In the molecular orbital description, what is the reason why the negative charge is not delocalised over all five atoms? What do the molecular orbitals for these anions look like?
If there are adjacent, parallel, overlapping p-orbitals in a molecule, the electrons will be delocalized over all such atoms. While the negative charge isn't delocalized over every atom in the conjugated system, it's more than you have shown. What you are failing to consider is:

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What's important to note about these resonances structures is that the net charge remains constant. These structures are consistent with both the reactivities of the enolate and $\alpha$,$\beta$-unsaturated carbonyl compounds. There negative charge is delocalized between the carbonyl oxygen and its $\alpha$-carbon, and the positive charge is delocalized between the carbonyl carbon (giving rise to 1,2-addition) and its $\beta$-carbon (giving rise to 1,4-addition).