1. Let the original number be represented as $N = 100a + 10b + c$, where $a, b,$ and $c$ are digits and $a \neq 0$ since $N$ is a positive integer.
2. According to the problem, when the leftmost digit $a$ is deleted, the resulting number is $10b + c$.
3. The problem states that this resulting number is $\frac{1}{29}$ of the original number. Therefore, we can write the equation:
$$10b + c = \frac{1}{29} (100a + 10b + c)$$
4. To eliminate the fraction, multiply both sides by 29:
$$29(10b + c) = 100a + 10b + c$$
5. Simplify and rearrange the equation:
$$290b + 29c = 100a + 10b + c$$
$$290b + 29c - 10b - c = 100a$$
$$280b + 28c = 100a$$
6. Factor out common terms:
$$28(10b + c) = 100a$$
$$25a = 70b + 7c$$
7. Since $25a$ must be a multiple of 7, $a$ must be a multiple of 7. The only single-digit multiples of 7 are 0 and 7. Since $a \neq 0$, we have $a = 7$.
8. Substitute $a = 7$ into the equation:
$$25 \cdot 7 = 70b + 7c$$
$$175 = 70b + 7c$$
9. Divide the entire equation by 7:
$$25 = 10b + c$$
10. We need to find single-digit values for $b$ and $c$ that satisfy this equation. By inspection, we find:
$$b = 2 \quad \text{and} \quad c = 5$$
11. Therefore, the original number is:
$$N = 100a + 10b + c = 100 \cdot 7 + 10 \cdot 2 + 5 = 700 + 20 + 5 = 725$$

The final answer is $\boxed{725}$.