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The natural density is $0$. This is due to the function $\sigma(n)/n$ possessing a continuous distribution, i.e. there exists a continuous function $\Phi:\mathbb{R}\to \mathbb{R}$ such that for all $a<b$ real numbers one has $$\lim_{x\to \infty}\frac{1}{x}\#\Big\{1\leq n \leq x: a<\frac{\sigma(n)}{n} \leq b\Big\}=\Phi(b)-\Phi(a).$$ Taking any $\epsilon>0$ and considering only $n>\frac{1}{\epsilon}$ one sees that the natural density of integers $n$ such that $\sigma(n)=2n-1$ is at most the density of integers $n$ with $$-\epsilon+2<\frac{\sigma(n)}{n}\leq 2$$ which equals $\Phi(2)-\Phi(2-\epsilon).$ Now since $\Phi$ is continuous we see that taking arbitrarily small $\epsilon$ the density of your integers $M$ is smaller than any positive number so it must vanish. EDIT:(references) According to this paper of Erdős the first published proof was given by Davenport in a 1934 paper not listed in Mathscinet but the method was identically the same as Schoenberg's earlier result that $\phi(n)/n$ has a continuous distribution.

The natural density is $0$. This is due to the function $\sigma(n)/n$ possessing a continuous distribution, i.e. there exists a continuous function $\Phi:\mathbb{R}\to \mathbb{R}$ such that for all $a<b$ real numbers one has $$\lim_{x\to \infty}\frac{1}{x}\#\Big\{1\leq n \leq x: a<\frac{\sigma(n)}{n} \leq b\Big\}=\Phi(b)-\Phi(a).$$ Fix any $k \in \mathbb{Z}$. Considering any $\epsilon>0$ we see that for $n>\frac{|k|}{\epsilon}$ the natural density of integers $n$ such that $\sigma(n)=2n+k$ is at most the density of integers $n$ with $$-\epsilon+2<\frac{\sigma(n)}{n}\leq 2$$ which equals $\Phi(2)-\Phi(2-\epsilon).$ Now since $\Phi$ is continuous we see that taking arbitrarily small $\epsilon$ the density of your integers $M$ is smaller than any positive number so it must vanish. Obviously for a specific choice of $k$ one can do better but the approach here does in fact provide density $0$ in very general situations.
EDIT:(references) According to this paper of Erdős the first published proof was given by Davenport in a 1934 paper not listed in Mathscinet but the method was identically the same as Schoenberg's earlier result that $\phi(n)/n$ has a continuous distribution.

The natural density is $0$. This is due to the function $\sigma(n)/n$ possessing a continuous distribution, i.e. there exists a continuous function $\Phi:\mathbb{R}\to \mathbb{R}$ such that for all $a<b$ real numbers one has $$\lim_{x\to \infty}\frac{1}{x}\#\Big\{1\leq n \leq x: a<\frac{\sigma(n)}{n} \leq b\Big\}=\Phi(b)-\Phi(a).$$ Taking any $\epsilon>0$ and considering only $n>\frac{1}{\epsilon}$ one sees that the natural density of integers $n$ such that $\sigma(n)=2n-1$ is at most the density of integers $n$ with $$-\epsilon+2<\frac{\sigma(n)}{n}\leq 2$$ which equals $\Phi(2)-\Phi(2-\epsilon).$ Now since $\Phi$ is continuous we see that taking arbitrarily small $\epsilon$ the density of your integers $M$ is smaller than any positive number so it must vanish.

The natural density is $0$. This is due to the function $\sigma(n)/n$ possessing a continuous distribution, i.e. there exists a continuous function $\Phi:\mathbb{R}\to \mathbb{R}$ such that for all $a<b$ real numbers one has $$\lim_{x\to \infty}\frac{1}{x}\#\Big\{1\leq n \leq x: a<\frac{\sigma(n)}{n} \leq b\Big\}=\Phi(b)-\Phi(a).$$ Taking any $\epsilon>0$ and considering only $n>\frac{1}{\epsilon}$ one sees that the natural density of integers $n$ such that $\sigma(n)=2n-1$ is at most the density of integers $n$ with $$-\epsilon+2<\frac{\sigma(n)}{n}\leq 2$$ which equals $\Phi(2)-\Phi(2-\epsilon).$ Now since $\Phi$ is continuous we see that taking arbitrarily small $\epsilon$ the density of your integers $M$ is smaller than any positive number so it must vanish. EDIT:(references) According to this paper of Erdős the first published proof was given by Davenport in a 1934 paper not listed in Mathscinet but the method was identically the same as Schoenberg's earlier result that $\phi(n)/n$ has a continuous distribution.

The natural density is $0$. This is due to the function $\sigma(n)/n$ possessing a continuous distribution, i.e. there exists a continuous function $\Phi:\mathbb{R}\to \mathbb{R}$ such that for all $a<b$ real numbers one has $$\lim_{x\to \infty}\frac{1}{x}\#\Big\{1\leq n \leq x: a<\frac{\sigma(n)}{n} \leq b\Big\}=\Phi(b)-\Phi(a).$$ Taking any $\epsilon>0$ and considering only $n>\frac{1}{\epsilon}$ one sees that the natural density of integers $n$ such that $\sigma(n)=2n-1$ is at most the density of integers $n$ with $$-\epsilon+\frac{1}{2}<\frac{\sigma(n)}{n}\leq \frac{1}{2}$$ which equals $\Phi(\frac{1}{2})-\Phi(\frac{1}{2}-\epsilon).$ Now since $\Phi$ is continuous we see that taking arbitrarily small $\epsilon$ the density of your integers $M$ is smaller than any positive number so it must vanish.

The natural density is $0$. This is due to the function $\sigma(n)/n$ possessing a continuous distribution, i.e. there exists a continuous function $\Phi:\mathbb{R}\to \mathbb{R}$ such that for all $a<b$ real numbers one has $$\lim_{x\to \infty}\frac{1}{x}\#\Big\{1\leq n \leq x: a<\frac{\sigma(n)}{n} \leq b\Big\}=\Phi(b)-\Phi(a).$$ Taking any $\epsilon>0$ and considering only $n>\frac{1}{\epsilon}$ one sees that the natural density of integers $n$ such that $\sigma(n)=2n-1$ is at most the density of integers $n$ with $$-\epsilon+2<\frac{\sigma(n)}{n}\leq 2$$ which equals $\Phi(2)-\Phi(2-\epsilon).$ Now since $\Phi$ is continuous we see that taking arbitrarily small $\epsilon$ the density of your integers $M$ is smaller than any positive number so it must vanish.