Solution:

A cousin $T$ is the image of an injection sending each $x \in S$ to $x - 1$ or $x + 1,$ landing outside $S.$ If $x - 1, x, x + 1 \in S$ then $x$ has nowhere to go, so every maximal block of consecutive elements of $S$ has size $1$ or $2.$ A double block $\{a, a+1\}$ is forced to map to $\{a - 1, a + 2\},$ while a singleton $\{a\}$ chooses $a - 1$ or $a + 1.$ Two blocks can fight over a value only when exactly one integer separates them, so group blocks into chains: consecutive blocks with gaps of exactly one. Within a chain the only consistent patterns are "the first $i$ blocks shift left and the rest shift right," since a block choosing right and its successor choosing left would collide; a double block acts as both left and right, forcing the switch to happen exactly at it. Hence a chain of $k$ singletons produces $k + 1$ distinct images, a chain containing one double produces exactly $1,$ and a chain with two doubles produces $0.$ Distinct patterns give distinct sets $T,$ and choices in different chains are independent, so the number of cousins is the product of $(k_i + 1)$ over the all-singleton chains.

We need $\prod (k_i + 1) = 4040 = 2^3 \cdot 5 \cdot 101$ while minimizing the element count $\sum k_i$ (chains with doubles only waste elements). Replacing a composite factor $f = gh$ with the two factors $g, h \ge 2$ strictly lowers the cost, because $(g - 1) + (h - 1) \lt gh - 1.$ So the optimum uses the prime factorization: $$\sum (f - 1) = 1 + 1 + 1 + 4 + 100 = 107,$$ realized by five chains of $1, 1, 1, 4, 100$ singletons — runs of every-other integer — placed far apart.

The least possible number of elements is $107.$