Call a k-digit positive integer a [i]hyperprime[/i] if all its segments consisting of $ 1, 2, ..., k$ consecutive digits are prime. Find all hyperprimes.
1. **Identify the digits of a hyperprime:**
   Since each digit of a hyperprime must be prime, the digits can only be $2, 3, 5,$ or $7$. This is because these are the only single-digit prime numbers.

2. **Single-digit hyperprimes:**
   Each of these digits as a one-digit number is prime:
   \[
   2, 3, 5, 7
   \]

3. **Two-digit hyperprimes:**
   For a two-digit number or higher, we need to ensure that all segments of the number are prime. Let's consider the possible combinations:
   - The digits $2$ and $5$ can only exist as the leftmost digit of the number because any segment starting with $2$ or $5$ must be prime.
   - The part of the number after the first digit must also be a hyperprime.

4. **Hyperprimes consisting of only $3$'s and $7$'s:**
   - We cannot have two consecutive $3$'s or two consecutive $7$'s, since neither $33$ nor $77$ is prime.
   - Therefore, the number must alternate $3$'s and $7$'s.
   - We also cannot have more than one $7$ because then the segment $737 = 11 \cdot 67$ is not prime.

   Thus, the possible hyperprimes using only $3$'s and $7$'s are:
   \[
   3, 7, 37, 73, 373
   \]

5. **Adding a $2$ or $5$ before the hyperprimes:**
   - $23$ and $53$ are both hyperprimes because $2$ and $3$, and $5$ and $3$ are prime.
   - Adding a $2$ or a $5$ to any of the other hyperprimes listed above would make the sum of its digits divisible by $3$ (so the new number itself will be divisible by $3$, and hence not prime).

6. **Verification:**
   - Check each number to ensure all segments are prime:
     \[
     2, 3, 5, 7, 23, 37, 53, 73, 373
     \]
   - All these numbers are indeed hyperprimes.

The final answer is $\boxed{2, 3, 5, 7, 23, 37, 53, 73, 373}$.