Let $P$ be a marvelous point of $ABC$ and let (without loss of generality) the area of $ABC$ be $27$. We first note that among the $27$ rays, there must always be $PA$, $PB$ and $PC$; otherwise not all the shapes determined by the rays will be triangles.

Each of the $27$ smaller triangles is contained within (or is equal to) one of the triangles $PAB$, $PBC$, $PCA$. Let $PAB$ contain $m$, and $PBC$ contain $n$ smaller triangles. Numbers $m$ and $n$ are elements of $\{1, 2, \dots, 25\}$ and they satisfy $m + n \le 26$.

We now consider the function which assigns the pair $(m, n)$ to each marvelous point of $ABC$. We claim that it is a bijection from the set of marvelous points to the set of pairs of positive integers $(m, n)$ which satisfy $m + n \le 26$.

We first show that for each pair $(m, n)$ we have at most one marvelous point. Let $P$ be a marvelous point assigned to $(m, n)$. This means that $PAB$ is divided into $m$ triangles, while $PBC$ is divided into $n$ triangles, each of whose areas equal $1$. The $m$ triangles contained within $PAB$ have a common altitude whose length equals $\frac{2m}{|AB|}$, so we know that point $P$ lies on the line $p$ which is parallel to $AB$, and whose distance to $AB$ equals $\frac{2m}{|AB|}$. Analogously, we infer that point $P$ lies on the line $q$ which is parallel to $BC$, and whose distance to $BC$ equals $\frac{2n}{|BC|}$. Since $p$ and $q$ are uniquely determined by $m$ and $n$, the fact that their intersection contains at most one point shows that there is at most one point to which the pair $(m, n)$ is assigned.

We now show that each pair $(m, n)$ is indeed assigned to a marvelous point. Given $m$ and $n$, consider the point $P$ defined as the intersection of lines $p$ and $q$, as in the previous paragraph. Drawing rays from point $P$, we can divide triangles $PAB$ and $PBC$ into $m$ and $n$ smaller triangles (of area equal to $1$), respectively. Finally, triangle $PAC$ has area $27 - m - n$, so we can divide it into $27 - m - n$ triangles of area $1$ by drawing rays originating in $P$. Thus, we can divide the triangle $ABC$ into $27$ smaller triangles of equal areas by drawing rays originating in $P$. By definition, this means that $P$ is a marvelous point to which the pair $(m, n)$ is assigned.

We have yet to determine the number of ways in which the pair $(m, n)$ can be chosen. For each $m \in \{1, 2, \dots, 25\}$, we can choose $n$ in exactly $26 - m$ ways. Therefore, the solution is
$$\sum_{m=1}^{25} (26 - m) = \frac{25 \cdot 26}{2} = 325.$$