The number $2021$ leaves a remainder of $11$ when divided by a positive integer. Find the smallest such integer.
1. We are given that the number \(2021\) leaves a remainder of \(11\) when divided by a positive integer \(d\). This can be written as:
   \[
   2021 \equiv 11 \pmod{d}
   \]
   This implies:
   \[
   2021 = kd + 11 \quad \text{for some integer } k
   \]
   Rearranging, we get:
   \[
   2021 - 11 = kd \implies 2010 = kd
   \]
   Therefore, \(d\) must be a divisor of \(2010\).

2. We need to find the smallest divisor of \(2010\) that is greater than \(11\). First, we factorize \(2010\):
   \[
   2010 = 2 \times 3 \times 5 \times 67
   \]
   The divisors of \(2010\) are: \(1, 2, 3, 5, 6, 10, 15, 30, 67, 134, 201, 335, 402, 670, 1005, 2010\).

3. From the list of divisors, the smallest divisor greater than \(11\) is \(15\).

Conclusion:
The smallest positive integer \(d\) such that \(2021\) leaves a remainder of \(11\) when divided by \(d\) is \(\boxed{15}\).