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If $n$ is even, then the answer is "yes". Take the direct product of a simple (infinite) group and ${\mathbb Z}/2^j{\mathbb Z}$. Every normal subgroup either is inside the finite cyclic group or contains the simple group. Total number is twice the number of normal subgroups of the cyclic group.If $n$ is odd, you would need to take a cyclic central extension of a simple group. That is also possible (the simple group can be, say, the Tarski monster, see our paper with Olshanskii and Osin on Lacunary hyperbolic groups in the arXiv).

Edit. If you want f.p. groups, look at the central extensions of the Thompson group $T$ described here.

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If $n$ is even, then the answer is "yes". Take the direct product of a simple (infinite) group and ${\mathbb Z}/2^j{\mathbb Z}$? . Every normal subgroup either is inside the finite cyclic group or contains the simple group. Total number is twice the number of normal subgroups of the cyclic group.If $n$ is odd, you would need to take a cyclic central extension of a simple group. That is also possible (the simple group can be, say, the Tarski monster, see our paper with Olshanskii and Osin on Lacunary hyperbolic groups in the arXiv).

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If $n$ is even, then the answer is "yes". Take the direct product of a simple (infinite) group and ${\mathbb Z}/2^j{\mathbb Z}$ ? Every normal subgroup either is inside the finite cyclic group or contains the simple group. Total number is twice the number of normal subgroups of the cyclic group.If $n$ is odd, you would need to take a cyclic central extension of a simple group. That is also possible (the simple group can be, say, the Tarski monster, see our paper with Olshanskii and Osin on Lacunary hyperbolic groups in the arXiv).