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If $S\subset\mathbb{N}$ has density $1$, then for any $m\in\mathbb{N}$ the density of $S\cap m\mathbb{N}$ is $1/m$, whence the density of $$S(m):=\{n\in\mathbb{N}:nm\in S\}$$ is $1$. Hence for any $m_1,\dotsc,m_k\in\mathbb{N}$, the density of $S(m_1)\cap\dotsb\cap S(m_k)$ is $1$. In particular, there exists $n\in\mathbb{N}$ such that $nm_1,\dotsc,nm_k\in S$. So any finite progression in $\mathbb{N}$ has a scaled version lying in $S$.