The number $N$ is the product of two primes. The sum of the positive divisors of $N$ that are less than $N$ is $2014$. Find $N$.
1. Let \( N = a \cdot b \), where \( a \) and \( b \) are prime numbers.
2. The positive divisors of \( N \) are \( 1, a, b, \) and \( N \).
3. The sum of all positive divisors of \( N \) is given by:
   \[
   1 + a + b + N
   \]
4. According to the problem, the sum of the positive divisors of \( N \) that are less than \( N \) is 2014. Therefore:
   \[
   1 + a + b = 2014
   \]
5. Rearrange the equation to find \( a + b \):
   \[
   a + b = 2014 - 1 = 2013
   \]
6. Since \( a \) and \( b \) are prime numbers, we need to find two primes that add up to 2013.
7. Checking the possible pairs, we find that \( a = 2 \) and \( b = 2011 \) (or vice versa) satisfy the condition, as both 2 and 2011 are prime numbers.
8. Therefore, the product \( N \) is:
   \[
   N = 2 \cdot 2011 = 4022
   \]

The final answer is \(\boxed{4022}\)