Count by apex: each vertex is the apex of $\lfloor \frac{n-1}{2} \rfloor$ isosceles triangles, but equilateral triangles get counted three times

Write $f(n)$ with a floor function and compute $f(n+1) - f(n)$ separately in each residue class of $n$ modulo $6$

Count isosceles triangles by apex. For a vertex $P$ of a regular $n$-gon, the isosceles triangles whose two equal sides meet at $P$ have their other two vertices symmetric about the diameter through $P,$ giving $\lfloor \frac{n-1}{2} \rfloor$ such pairs. Summing over all $n$ vertices counts each non-equilateral isosceles triangle once (it has one apex) and each equilateral triangle three times; equilateral triangles exist exactly when $n$ is a multiple of $3,$ and then there are $\frac{n}{3}$ of them. Hence $f(n) = n \lfloor \frac{n-1}{2} \rfloor,$ minus $\frac{2n}{3}$ when $n$ is a multiple of $3.$

Writing $n = 6k + j$ and computing $f(n+1) - f(n)$ in each residue class gives $13k$ for $j = 0,$ $3k$ for $j = 1,$ $5k + 1$ for $j = 2,$ $7k + 3$ for $j = 3,$ $9k + 6$ for $j = 4,$ and $-(k + 2)$ for $j = 5.$ Setting each equal to $78:$ $13k = 78$ gives $k = 6,$ $n = 36;$ $3k = 78$ gives $k = 26,$ $n = 157;$ $9k + 6 = 78$ gives $k = 8,$ $n = 52;$ and the other three cases have no positive integer solutions.

The sum of all such $n$ is $36 + 157 + 52 = 245.$