Throughout the solution, we denote by $[a, b]$ the set $\{a, a+1, \ldots, b\}$. We say that $\{a, b, c\}$ is an obtuse triple if $a, b, c$ are the sides of some obtuse triangle. We prove by induction on $n$ that there exists a partition of $[2,3n+1]$ into $n$ obtuse triples $A_i$ $(2 \leq i \leq n+1)$ having the form $A_i=\{i, a_i, b_i\}$. For the base case $n=1$, one can simply set $A_2=\{2,3,4\}$. For the induction step, we need the following simple lemma.

Lemma. Suppose that the numbers $a(c-b)(c+b)>a^2$.

Now we turn to the induction step. Let $n>1$ and put $t=\lfloor n / 2\rfloor \cdot \frac{n}{2} \cdot \frac{9n}{2} \geq (n+1)^2 \geq i^2$, so this triangle is obtuse. The proof is completed.