Solution:

Let's first count the number of such triangles with perimeter equal to $p$. By Stars and Bars, there are $\binom{p}{2} \approx \frac{p^{2}}{2}$ ordered triples of positive integers that sum to $p$. Additionally, note that only about a quarter of them satisfy the triangle inequality, we have only $\frac{p^{2}}{8}$ possible triples. Dividing by $3!$ gives us approximately $\frac{p^{2}}{48}$ nondegenerate triangles with perimeter $p$. Summing this from $p=1$ to $n$ gives us approximately $\frac{n^{3}}{144}$ triangles with perimeter at most $n$.

Now, note that there are two "extremes" for our triangles. One extreme is a triangle that is very close to a line. In that case, we have that the maximum perimeter is $200 \sqrt{2} \approx 283$. In the other extreme, we have a triangle that is very close to an equilateral triangle, in which case we have the maximum perimeter is $3 \cdot \frac{100}{\cos 15^{\circ}} \approx 311$. Thus, as a compromise between these extremes, we can plug in $n=300$ to get a value of $\frac{300^{3}}{144}=187500$, which would have earned 13 points.