Partitioning the naturals into an infinite number of large sets Is it possible to partition the positive integers into an infinite number of disjoint large sets ?
 A: Similar, a squarefree number is $1,2,3,5,6,7,10,11,13,15,\ldots$ so let your disjoint sets be $m n^2$ for squarefree $m.$ These sets are called squareclasses, I like to write it as one word. The equivalence relation is that two numbers are equivalent if and only if their ratio is a rational square. As we are dealing with integers, that is the same as saying that two numbers are equivalent if and only if their product is a square.
This simple idea underlies a good deal of integer coefficient quadratic form theory. In particular, we speak of squareclasses in the $p$-adic numbers $\mathbb Q_p$ and   the $p$-adic integers $\mathbb Z_p.$ See, for example, Cassels, Rational Quadratic Forms. 
A: Let $N_k$ be the integers with exactly $k$ prime factors (counting with multiplicity).  The $N_k$ for $k \ge 1$ partition the integers and are all infinite, if you just agree to stow  $1$ somewhere.
A: Yes. $D=\{1,3,5,7,9,\dots\}$, the set of all odd numbers, is a large set; so is $2D=\{2,6,10,14,18,\dots\}$; so are $4D,8D,16D$, etc.
A: I understand that the question has already been answered, and answered well, but here is another very quick, but slightly different argument:
The usual proof that $\sum \frac{1}{n}$ diverges uses the fact that there are an infinite number of disjoint finite blocks of numbers whose reciprocals add up to more than $\frac{1}{2}$:
$$\{1\} , \{2,3\}, \{4,5,6,7\}, \{8,9,10,\ldots,15\}, \ldots$$
Call these blocks $B_{1}, B_{2},\ldots$
It is obvious that any set of integers that contains an infinite number of these blocks must be a large set.  So, just partition $\mathbb{N}$ into an infinite number of infinite sets $\{A_{i}\vert i\in \mathbb{N}\}$, and define the large set $L_{i}$ as
$$L_{i}=\bigcup_{x\in A_{i}}B_{i}.$$ 
