How many ordered triples $(a, b, c)$ of odd positive integers satisfy $a + b + c = 25?$
1. Since \(a, b, c\) are odd positive integers, we can express them in the form \(a = 2x + 1\), \(b = 2y + 1\), and \(c = 2z + 1\) where \(x, y, z\) are non-negative integers. This is because any odd integer can be written as \(2k + 1\) for some integer \(k\).

2. Substitute these expressions into the equation \(a + b + c = 25\):
   \[
   (2x + 1) + (2y + 1) + (2z + 1) = 25
   \]

3. Simplify the equation:
   \[
   2x + 2y + 2z + 3 = 25
   \]

4. Subtract 3 from both sides:
   \[
   2x + 2y + 2z = 22
   \]

5. Divide the entire equation by 2:
   \[
   x + y + z = 11
   \]

6. We need to find the number of non-negative integer solutions to the equation \(x + y + z = 11\). This is a classic problem of finding the number of solutions to an equation in the form of \(x_1 + x_2 + \cdots + x_k = n\), which can be solved using the "stars and bars" theorem.

7. According to the "stars and bars" theorem, the number of non-negative integer solutions to the equation \(x_1 + x_2 + \cdots + x_k = n\) is given by:
   \[
   \binom{n + k - 1}{k - 1}
   \]
   Here, \(n = 11\) and \(k = 3\) (since we have three variables \(x, y, z\)).

8. Substitute \(n = 11\) and \(k = 3\) into the formula:
   \[
   \binom{11 + 3 - 1}{3 - 1} = \binom{13}{2}
   \]

9. Calculate \(\binom{13}{2}\):
   \[
   \binom{13}{2} = \frac{13 \times 12}{2 \times 1} = 78
   \]

Conclusion:
The number of ordered triples \((a, b, c)\) of odd positive integers that satisfy \(a + b + c = 25\) is \(\boxed{78}\).