, 2022

The answer is $\binom{50}{3}$ minus the number of choices of three rods that violate the triangle inequality. The number to be subtracted is the size of the set
$$S = \{(a, b, c) : 1 \le a < b < c \le 99,\ a + b < c,\ a, b, c \text{ are odd numbers}\}.$$
Note that for $(a, b, c) \in S$, $a+b$ is an even number between $4$ and $98$ (inclusive). For each such even number, we can easily count the number of such pairs $(a, b)$, and for each such $(a, b)$ we can easily count the number of choices of $c$ for which $(a, b, c) \in S$.

As an example, if we fix $a + b = 20$, there would be $5$ choices for $(a, b)$, namely, $(1, 19)$, $(3, 17)$, $(5, 15)$, $(7, 13)$ and $(9, 11)$. For each of these $5$ pairs of $(a, b)$, there are $40$ choices of $c$ for which $(a, b, c) \in S$, namely, $21, 23, 25, \ldots, 97, 99$.
In the same way, we can see that for each even number $2k$ where $k \in \{2, 3, 4, \dots, 48, 49\}$, there would be $\lfloor \frac{1}{2}k \rfloor$ choices of $(a, b)$ for which $a + b = 2k$, and for each such $(a, b)$ there would be $50-k$ choices of $c$ for which $(a, b, c) \in S$. This can be summarized by the following table: