There exist two distinct positive integers, both of which are divisors of $10^{10}$, with sum equal to $157$.  What are they?
1. We start by noting that the two positive integers must be of different parities because their sum is odd. This is a fundamental property of integers: the sum of two even numbers or two odd numbers is even, while the sum of an even number and an odd number is odd.

2. Next, we recognize that the two integers must be divisors of \(10^{10}\). We can express \(10^{10}\) as:
   \[
   10^{10} = (2 \cdot 5)^{10} = 2^{10} \cdot 5^{10}
   \]
   Therefore, any divisor of \(10^{10}\) can be written in the form \(2^a \cdot 5^b\) where \(0 \leq a \leq 10\) and \(0 \leq b \leq 10\).

3. Since the sum of the two integers is 157, we denote the two integers as \(2^a \cdot 5^b\) and \(2^c \cdot 5^d\) such that:
   \[
   2^a \cdot 5^b + 2^c \cdot 5^d = 157
   \]

4. Given that the sum is odd, one of the integers must be even and the other must be odd. This implies that one of the integers must be a power of 2 (since powers of 2 are even) and the other must be a power of 5 (since powers of 5 are odd).

5. We now test possible values for \(a\) and \(b\) to find pairs that sum to 157. We start by considering the possible values for \(5^b\):
   - \(5^0 = 1\)
   - \(5^1 = 5\)
   - \(5^2 = 25\)
   - \(5^3 = 125\)
   - \(5^4 = 625\) (too large since \(2^a\) must be positive and less than 157)

6. We test each value of \(5^b\) to see if there exists a corresponding \(2^a\) such that their sum is 157:
   - For \(5^0 = 1\):
     \[
     2^a + 1 = 157 \implies 2^a = 156 \quad (\text{not a power of 2})
     \]
   - For \(5^1 = 5\):
     \[
     2^a + 5 = 157 \implies 2^a = 152 \quad (\text{not a power of 2})
     \]
   - For \(5^2 = 25\):
     \[
     2^a + 25 = 157 \implies 2^a = 132 \quad (\text{not a power of 2})
     \]
   - For \(5^3 = 125\):
     \[
     2^a + 125 = 157 \implies 2^a = 32 \quad (\text{and } 32 = 2^5)
     \]

7. We find that \(2^5 = 32\) and \(5^3 = 125\) are the two integers that sum to 157 and are both divisors of \(10^{10}\).

The final answer is \(\boxed{32, 125}\).