Let us denote $B=\mathbb N\setminus N$, i.e., the set of consecutive-summable numbers.

There already is an answer precisely characterizing the sets $N$ and $B$. (Which is much better result than just estimating the density.) If we are only interested in the density, another way to get that it is equal to zero is to notice the following.

If we fix a $k\in\mathbb N$, then we get the numbers of the form $$n=km+\sum_{i=0}^k k = km + T_k$$ where $T_k$ is the $k$-th triangular number. So we see that with finitely many exceptions $B$ contains the arithmetic progression $k\mathbb N+a$, where $a=T_k \bmod k$.

That means that $N$ is almost (i.e., with finitely many exceptions) contained in the union of the remaining $(k-1)$ congruence classes modulo $k$.

If we take any coprime $k_1,\ldots,k_n$, then we have that $N$ is almost contained in the union of corresponding congruence classes modulo $k_1\cdots k_n$. (Here we are using the Chinese remainder theorem which gives us the correspondence between the system of remainder modulo $k_1,\dots,k_n$ and a single congruence class modulo $k_1\cdots k_n$.) This gives us an estimate for the upper density $$\operatorname{ud}(N) \le \frac{(k_1-1)\cdot (k_2-1) \cdot (k_n-1)}{k_1\cdot k_2 \cdot k_n} = \prod_{j=1}^n \left(1-\frac1{k_j}\right).$$

In particular, if we take $k_n=p_n$ to be the $n$-th prime number, we get that $$\operatorname{ud}(N) \le \prod_{j=1}^n \left(1-\frac1{k_j}\right)$$ and since this is true for every $n$ we have $$\operatorname{ud}(N) \le \prod_{j=1}^\infty \left(1-\frac1{p_j}\right)=0.$$

You may notice that this argument would work for any set which can be related in a similar way to the union of distinct congruence classes if we have $\sum \frac1{k_j}=\infty$.

Let us denote $B=\mathbb N\setminus N$, i.e., the set of consecutive-summable numbers.

There already is an answer precisely characterizing the sets $N$ and $B$. (Which is much better result than just estimating the density.) If we are only interested in the density, another way to get that it is equal to zero is to notice the following.

If we fix a $k\in\mathbb N$, then we get the numbers of the form $$n=km+\sum_{i=0}^k k = km + T_k$$ where $T_k$ is the $k$-th triangular number. So we see that with finitely many exceptions $B$ contains the arithmetic progression $k\mathbb N+a$, where $a=T_k \bmod k$.

That means that $N$ is almost (i.e., with finitely many exceptions) contained in the union of the remaining $(k-1)$ congruence classes modulo $k$.

If we take any coprime $k_1,\ldots,k_n$, then we have that $N$ is almost contained in the union of corresponding congruence classes modulo $k_1\cdots k_n$. (Here we are using the Chinese remainder theorem which gives us the correspondence between the system of remainder modulo $k_1,\dots,k_n$ and a single congruence class modulo $k_1\cdots k_n$.) This gives us an estimate for the upper density $$\operatorname{ud}(N) \le \frac{(k_1-1)\cdot (k_2-1) \cdot (k_n-1)}{k_1\cdot k_2 \cdot k_n} = \prod_{j=1}^n \left(1-\frac1{k_j}\right).$$

In particular, if we take $k_n=p_n$ to be the $n$-th prime number, we get that $$\operatorname{ud}(N) \le \prod_{j=1}^n \left(1-\frac1{p_j}\right)$$ and since this is true for every $n$ we have $$\operatorname{ud}(N) \le \prod_{j=1}^\infty \left(1-\frac1{p_j}\right)=0.$$

You may notice that this argument would work for any set which can be related in a similar way to the union of distinct congruence classes if we have $\sum \frac1{k_j}=\infty$.