So many triangles!

Let's try some simple cases first so we can get something inspired for the actual proof.

If the perpendicular sides are parallel to the coordinate axes, we find that we can generate all multiples of 3.

Now we try if the vertices are $(0,0)$ , $(1,1)$ and $(1, -1)$ . This gives us 4 lattice points, and if we continue doing triangles of the form $(0,0)$ , $(m,m)$ and $(m,-m)$ , we find we can generate all multiples of 4.

Now we will try the proof and see fi we can get something like the above results.

For sake of convenience, we can assume the apex of the triangle is $(0,0)$ , and the other vertices are $(m,n)$ and $(n, -m)$ . Note that the number of lattice points between the origin and $(m,n)$ is $\text{gcd}(m,n) + 1$ . The same holds true for the number of lattice points between the origin and $(n,-m)$ . To count the number of lattice points on the hypotenuse, we translate the side so one of the vertices (assume it's $(n,-m)$ ) is now at $(0,0)$ . Thus, the other vertex must be at $(m-n,m+n)$ , so the number of lattice points on the hypotenuse must be $\text{gcd}(m+n,m-n) + 1$ .

Let $k=\text{gcd}(m,n)$ . We must have that $k \mid m$ and $k \mid n$ , so $k \mid m+n, m-n$ . Thus, $k \mid \text{gcd}(m+n,m-n)$ .

Let $l=\text{gcd}(m+n,m-n)$ . We must have $l \mid m+n$ and $l \mid m-n$ , so $l \mid 2m, 2n$ . Thus, either $k=l$ if $l$ is odd or $2k=l$ if $l$ is even. Note that $l$ is odd if $m$ and $n$ have different parities and $l$ is even if $m$ and $n$ have the same parity.

Thus, if we $m$ and $n$ have different parity, we get there are $3l$ lattice points on the triangle (thus generating all multiples of 3), and if $m$ and $n$ have the same parity, we get there are $4l$ lattice points on the triangle (thus generating all multiples of 4).

We already have a pair of constructions satisfying these two properties, so we are done!

Now all that remains is to count the number of positive integers which satisfy. By PIE, we have

$\left \lfloor \dfrac{2016}{3} \right \rfloor+ \left \lfloor \dfrac{2016}{4} \right \rfloor - \left \lfloor \dfrac{2016}{12} \right \rfloor = \boxed{1008}$

Therefore, our answer is $\boxed{1008}$ .