One trivial case is to have $16$ lines, no two of which are parallel. You could also have $18$ lines, divided into six groups of $3$ lines each such that every group contains only parallel lines. You could also have $20$ lines, divided into four groups of $5$ lines each such that every group contains only parallel lines. Finally, you could have $30$ lines divided into $2$ groups of $15$ lines each such that both groups contain parallel lines. These are the only possibilities because the only divisors of $15$ are $1,3,5,15$ and for each divisor $d$, we can have $\frac{15(d+1)}{d}$ lines divided into $\frac{15}{d}+1$ groups of $d$ parallel lines.

Define $a \sim b$ if $a$ is parallel to $b.$ Parallelity is an equivalence relation, and splits the set of lines into equivalence classes; suppose they have sizes $a_1 \ge a_2 \ge \dots \ge a_n > 0.$ Let $S=a_1+\dots+a_n.$ We have $S-a_i = 15$ for all $i,$ so $a_i = S-15$ for all $i$ and consequently $n(S-15) = \sum_i a_i = S,$ which implies $n = \frac{S}{S-15} = 1+\frac{15}{S-15}.$ Consequently, $(S-15)|15 \Rightarrow S = 16, 18, 20, 30 \Rightarrow n = 16, 6, 4, 2.$ These possibilities correspond to $16$ groups of $1$ line, $6$ groups of $3$ lines, $4$ groups of $5$ lines, and $2$ groups of $15$ lines, all of which work.