Find the smallest positive integer $ n$ such that $ 107n$ has the same last two digits as $ n$.
1. We need to find the smallest positive integer \( n \) such that \( 107n \) has the same last two digits as \( n \). This can be expressed mathematically as:
   \[
   107n \equiv n \pmod{100}
   \]

2. Simplify the congruence:
   \[
   107n \equiv n \pmod{100}
   \]
   Subtract \( n \) from both sides:
   \[
   107n - n \equiv 0 \pmod{100}
   \]
   \[
   106n \equiv 0 \pmod{100}
   \]

3. This implies that \( 106n \) is divisible by 100:
   \[
   106n = 100k \quad \text{for some integer } k
   \]

4. Simplify the equation:
   \[
   106n \equiv 0 \pmod{100}
   \]
   This can be rewritten as:
   \[
   6n \equiv 0 \pmod{100}
   \]

5. For \( 6n \equiv 0 \pmod{100} \), \( n \) must be such that \( 6n \) is a multiple of 100. Therefore, \( n \) must be a multiple of \( \frac{100}{6} \):
   \[
   n = \frac{100k}{6} = \frac{50k}{3}
   \]

6. Since \( n \) must be an integer, \( k \) must be a multiple of 3. The smallest positive integer \( k \) that satisfies this is \( k = 3 \):
   \[
   n = \frac{50 \cdot 3}{3} = 50
   \]

7. Therefore, the smallest positive integer \( n \) such that \( 107n \) has the same last two digits as \( n \) is:
   \[
   \boxed{50}
   \]