Compute the number of positive integer divisors of $2121$ with a units digit of $1$.


[i]2021 CCA Math Bonanza Individual Round #1[/i]
1. First, we need to find the prime factorization of \( 2121 \):
   \[
   2121 = 3 \times 7 \times 101
   \]
   This tells us that \( 2121 \) is the product of the primes \( 3 \), \( 7 \), and \( 101 \).

2. Next, we need to identify the divisors of \( 2121 \) that have a units digit of \( 1 \). To do this, we will list all possible divisors and check their units digits.

3. The divisors of \( 2121 \) are formed by taking products of the prime factors \( 3 \), \( 7 \), and \( 101 \). The complete list of divisors is:
   \[
   1, 3, 7, 21, 101, 303, 707, 2121
   \]

4. Now, we filter this list to find the divisors with a units digit of \( 1 \):
   \[
   1, 21, 101, 2121
   \]

5. We count the number of divisors with a units digit of \( 1 \):
   \[
   \text{Number of divisors with units digit } 1 = 4
   \]

The final answer is \(\boxed{4}\).