Consider the four cyclic species labeled (I) to (IV) in the given...

According to Huckel's rule, a species is aromatic if it is cyclic, planar, completely conjugated, and has $(4n + 2)$ $\pi$ electrons, where $n = 0, 1, 2, \dots$

For species (I), pyrrole: It is a cyclic, planar molecule. The nitrogen atom is $sp^2$ hybridized and its lone pair resides in a p-orbital, participating in conjugation. Total $\pi$ electrons = $4$ (from two double bonds) $+ 2$ (from lone pair) $= 6 \pi$ electrons. Since $6 = 4(1) + 2$, it follows Huckel's rule and is aromatic.

For species (II), cyclopropenyl anion: It is cyclic and planar. Total $\pi$ electrons = $2$ (from double bond) $+ 2$ (from negative charge) $= 4 \pi$ electrons. Since it has $4n$ $\pi$ electrons ($n=1$), it is antiaromatic.

For species (III), tropylium cation: It is a cyclic, planar molecule with complete conjugation due to the empty p-orbital on the positively charged carbon. Total $\pi$ electrons = $6$ (from three double bonds). Since $6 = 4(1) + 2$, it follows Huckel's rule and is aromatic.

For species (IV), cyclopentadiene: It contains an $sp^3$ hybridized carbon atom ($\text{CH}_2$), which breaks the continuous conjugation of the ring. Therefore, it is non-aromatic.

Thus, only (I) and (III) are aromatic.

Answer: I and III only