, 2016

The maximum number of acute-angled triangles is $2729148240$.
Let $n = 2015$, so that there are $2n + 1$ lines. We fix one of the lines $\ell$ and place it as the $x$-axis of the coordinate plane. Then the other $2n$ lines can be partitioned into two groups, one consisting of those lines with positive slopes and one consisting of those lines with negative slopes. Let $a$ and $b$ be the sizes of these two groups respectively.

Observe that every pair of lines belonging to the same group form an obtuse triangle with $\ell$. Therefore, the number of obtuse triangles having $\ell$ as a sideline that is adjacent to the obtuse angle is
$$\binom{a}{2} + \binom{b}{2} \ge 2\binom{n}{2} = n(n-1)$$
by Jensen's inequality, since the binomial function $\binom{x}{2}$ is convex.

By considering all choices of $\ell$, each obtuse triangle is counted twice. Therefore, the number of acute-angled triangles is at most
$$\binom{2n+1}{3} - \frac{(2n+1)n(n-1)}{2} = \frac{n(n+1)(2n+1)}{6}.$$
This bound is attainable. Indeed, consider the sidelines of a regular polygon with $2n + 1$ sides. Then for every line $\ell$ being the $x$-axis, the number of lines with positive slopes is equal to the number of lines with negative slopes by symmetry. Therefore, equality of the above deduction holds.

When $n = 2015$, the answer is $\frac{2015 \times 2016 \times 4031}{6} = 2729148240$.