Find the least positive integer such that when its leftmost digit is deleted, the resulting integer is $\frac{1}{29}$ of the original integer.
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 \(\boxed{725}\).