Counterexamples showing natural density is not a measure Natural density $d$ measures how large a subset of natural numbers is, as defined here.
There are some examples showing natural density is not countably summable. For instance, 
$$
0=\sum_{k=0}^\infty d(\{k\}) \neq d(\bigcup_{k=0}^\infty\{k\})=d(\mathbb{N})=1.
$$
However, all those examples I have seen considers sets with zero density.
Question: Is there a counterexample (showing that natural density is not countably summable) considers only the sets with density strictly between 0 and 1? That is to find $\sum_{k=0}^\infty d(A_k) \neq d(\bigcup_{k=0}^\infty A_k)$, where all $A_k$ are disjoint and $0<d(A_k)<1$ for all $k$ and $d(\bigcup_{k=0}^\infty A_k)<1$.
In other words, can $d$ be a measure if we restrict its domain to all the sets that have density strictly between 0 and 1 and $\mathbb{N}$ and $\emptyset$?
 A: It is relatively easy to find an example of disjoint sets that have density zero such that their union is the whole $\mathbb N$.
You can simply take singletons. If you want less trivial example, you can take1 $B_n=\{k!+n; k\in\mathbb N, k!+n<(k+1)!\}$ for $n=0,1,2\dots$.
Now let us take a decomposition of some set $C$ with $d(C)\in(0,1)$ into disjoint sets $C_k$ such that each of them have positive density and $d(C)=\sum d(C_k)$.2
Notice that if both $B$ and $C$ have density and $d(B)=0$, then 
$$d(C\cup B)=d(C)$$
simply because $d(C\cup B) = d(C\cup (B\setminus S))=d(C)+d(B\setminus C) = d(C)+0$.
So if you now take3 
$$A_k=C_k \cup \left( B_k \setminus \bigcup_{j<k} C_j\right)$$
then you have sets such that $d(A_k)=d(C_k)$ and the union $\bigcup\limits_{k=0}^\infty A_k$ is the whole set $\mathbb N$ and $\sum\limits_{k=0}^\infty d(A_k)=\sum\limits_{k=0}^\infty d(C_k)=d(C)$. This implies that
$$\sum_{k=0}^\infty d(A_k) \ne d\left(\bigcup\limits_{k=0}^\infty A_k\right)$$
since $d(C)<d(\mathbb N)=1$.
If you insist on union not having full density, you can simply rescale the whole situation by factor two.

1This means that we take $B_0=\{1!,2!,3!,4!,\dots\}$. Into the set $B_1$ we give the numbers of the form $k!+1$, but we omit those ones that already are in $B_0$. So we have $B_1=\{2!+1,3!+1,4!+1,\dots\}$. And you can continue in this manner. (For example, $B_4$ does contains neither $1!+5=6=2!+3\in B_3$ nor $2!+4=6\in B_1$, but it does contain $3!+4=10$.)
2We can take $C=2\mathbb N+1$ and $C_k=2^{2k+2}\mathbb N+2^{2k+1}+1$. 
I.e., $C_0=4\mathbb N+1$, $C_1=8\mathbb N+3$, $C_2=16\mathbb N+7$, etc.
3Here we basically take $C_k\cup B_k$, but to make them disjoint we have to remove the elements contained in the previous sets. This does not change density, since we only remove a set with zero density

Remark. If you are interested in the fact that at least set with density zero has some properties similar to $\sigma$-additivity, useful keywords to look for might be P-ideal, P-filter, property AP or property AP0 . 
