We know that $201$ and $9$ give the same remainder when divided by $24$. What is the smallest positive integer $k$ such that $201+k$ and $9+k$ give the same remainder when divided by $24+k$?

[i]2020 CCA Math Bonanza Lightning Round #2.1[/i]
1. We start with the given condition that \(201\) and \(9\) give the same remainder when divided by \(24\). This can be written as:
   \[
   201 \equiv 9 \pmod{24}
   \]
   Simplifying this, we get:
   \[
   201 - 9 = 192 \equiv 0 \pmod{24}
   \]
   This confirms that \(192\) is divisible by \(24\).

2. We need to find the smallest positive integer \(k\) such that \(201 + k\) and \(9 + k\) give the same remainder when divided by \(24 + k\). This can be written as:
   \[
   201 + k \equiv 9 + k \pmod{24 + k}
   \]
   Simplifying this, we get:
   \[
   201 + k - (9 + k) \equiv 0 \pmod{24 + k}
   \]
   \[
   192 \equiv 0 \pmod{24 + k}
   \]
   This means that \(24 + k\) must be a divisor of \(192\).

3. We factorize \(192\) to find its divisors:
   \[
   192 = 2^6 \cdot 3
   \]
   The divisors of \(192\) are \(1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 64, 96, 192\).

4. We need the smallest positive integer \(k\) such that \(24 + k\) is one of these divisors. We start with the smallest divisor greater than \(24\):
   \[
   24 + k = 32 \implies k = 32 - 24 = 8
   \]

Thus, the smallest positive integer \(k\) is \(8\).

The final answer is \(\boxed{8}\).