I'm not sure I could call any chord "unresolvable", though I'd have to know the context of the conversation.
However, augmented chords (e.g. C-E-G♯), as well as diminished 7th chords (e.g. C-E♭-G♭-B♭♭)
-- both of which have been mentioned in other answers -- share a common trait: they are, in some sense, symmetric. Augmented chords are created by stacking three major thirds, and diminshed 7ths are likewise created by stacking four minor thirds, both of which add to an octave. While neither chord is "unresolvable", these two types of symmetric chords could be said to be "ambiguously resolvable", in that they look the same from three or four different keys (respectively), and can therefore be plausibly resolved to any one of several possible distant keys. As a result, romantic composers will often use them as a sort of "turn-stile" to modulate between unrelated keys.
In fact, the Tristan Chord, mentioned in yet another existing answer, is also somewhat symmetric (at least after the G♯ resolves to the A). It consists of two major thirds (F-A and B-D♯), separated by a tritone (a symmetric interval). As such, you could almost envision it resolving to a Bb7 instead of an E7 (if it weren't for those chromatic passing tones that help to indicate a clear direction to the line).
The takeaway point is that, in music, symmetry leads to a certain ambiguity of tonal center, and therefore can weaken the sense of needing any specific resolution. Such a chord still needs a resolution, but it has several options, and so does not necessarily need any specific resolution.