Paul starts at $1$ and counts by threes: $1, 4, 7, 10, ... $. At the same time and at the same speed, Penny counts backwards from $2017$ by fives: $2017, 2012, 2007, 2002,...$ . Find the one number that both Paul and Penny count at the same time.
1. Let's denote the sequence that Paul counts by as \( a_n \). Since Paul starts at 1 and counts by threes, the sequence can be written as:
   \[
   a_n = 1 + 3(n-1) = 3n - 2
   \]
   where \( n \) is the number of terms.

2. Similarly, let's denote the sequence that Penny counts by as \( b_m \). Since Penny starts at 2017 and counts backwards by fives, the sequence can be written as:
   \[
   b_m = 2017 - 5(m-1) = 2022 - 5m
   \]
   where \( m \) is the number of terms.

3. We need to find the common term in both sequences, i.e., we need to solve for \( n \) and \( m \) such that:
   \[
   3n - 2 = 2022 - 5m
   \]

4. Rearrange the equation to isolate terms involving \( n \) and \( m \):
   \[
   3n - 2 = 2022 - 5m \implies 3n + 5m = 2024
   \]

5. We need to find integer solutions for \( n \) and \( m \). To do this, we can express \( n \) in terms of \( m \):
   \[
   3n = 2024 - 5m \implies n = \frac{2024 - 5m}{3}
   \]

6. For \( n \) to be an integer, \( 2024 - 5m \) must be divisible by 3. We check the divisibility condition:
   \[
   2024 \equiv 2 \pmod{3} \quad \text{and} \quad 5 \equiv 2 \pmod{3}
   \]
   \[
   2024 - 5m \equiv 2 - 2m \pmod{3}
   \]
   \[
   2 - 2m \equiv 0 \pmod{3} \implies 2(1 - m) \equiv 0 \pmod{3} \implies 1 - m \equiv 0 \pmod{3} \implies m \equiv 1 \pmod{3}
   \]

7. Therefore, \( m \) can be written as:
   \[
   m = 3k + 1 \quad \text{for some integer } k
   \]

8. Substitute \( m = 3k + 1 \) back into the equation for \( n \):
   \[
   n = \frac{2024 - 5(3k + 1)}{3} = \frac{2024 - 15k - 5}{3} = \frac{2019 - 15k}{3} = 673 - 5k
   \]

9. We need \( n \) to be a positive integer, so:
   \[
   673 - 5k > 0 \implies 673 > 5k \implies k < \frac{673}{5} \implies k < 134.6
   \]
   Thus, \( k \) can be any integer from 0 to 134.

10. For the smallest \( k \), \( k = 0 \):
    \[
    n = 673 - 5(0) = 673
    \]
    \[
    m = 3(0) + 1 = 1
    \]

11. Substitute \( n = 673 \) back into Paul's sequence:
    \[
    a_{673} = 3(673) - 2 = 2019 - 2 = 2017
    \]

12. Substitute \( m = 1 \) back into Penny's sequence:
    \[
    b_{1} = 2022 - 5(1) = 2022 - 5 = 2017
    \]

Thus, the common number that both Paul and Penny count at the same time is \( 2017 \).

The final answer is \(\boxed{2017}\).