How to prove that $ \sum_{n \in \mathbb{N} } | \frac{\sin( n)}{n} | $ diverges?

Hint: $|\sin n|\geqslant \sin^2 n$, and use the convergence of $\sum_{n=1}^{+\infty}\frac{\cos(\color{red}2n)}n$ and the divergence of harmonic series.

We have $\cos(2n)=2\cos^2n-1$, and $$\sin^2n=1-\cos^2n=1-\frac{\cos(2n)+1}2=\frac{1-\cos(2n)}2.$$

The idea in your comment leads to one approach:

For each positive integer $k$, the interval $\bigl[k\pi+{\pi\over 6}, (k+1)\pi-{\pi\over 6}\bigr]$ has length exceeding $1$ and thus contains an integer $n_k$. We then have ${|\sin (n_k)|\over n_k} \ge {\sin({\pi\over 6})\over (k+1)\pi}$. So, from the Comparison test, it follows that $\sum\limits_{k=1}^\infty {|\sin(n_k)|\over n_k}$ diverges; whence $\sum\limits_{n=1}^\infty {|\sin(n )|\over n}$ diverges (by the Comparison test again).