Let $ P_1$ be a regular $ r$-gon and $ P_2$ be a regular $ s$-gon $ (r\geq s\geq 3)$ such that each interior angle of $ P_1$ is $ \frac {59}{58}$ as large as each interior angle of $ P_2$. What's the largest possible value of $ s$?
1. The interior angle of a regular $r$-gon is given by:
   \[
   \text{Interior angle of } P_1 = \frac{180(r-2)}{r}
   \]
   
2. The interior angle of a regular $s$-gon is given by:
   \[
   \text{Interior angle of } P_2 = \frac{180(s-2)}{s}
   \]

3. According to the problem, each interior angle of $P_1$ is $\frac{59}{58}$ times as large as each interior angle of $P_2$. Therefore, we set up the equation:
   \[
   \frac{59}{58} \cdot \frac{180(s-2)}{s} = \frac{180(r-2)}{r}
   \]

4. Simplify the equation:
   \[
   \frac{59 \cdot 180(s-2)}{58s} = \frac{180(r-2)}{r}
   \]

5. Cancel out the common factor of 180 from both sides:
   \[
   \frac{59(s-2)}{58s} = \frac{r-2}{r}
   \]

6. Cross-multiply to eliminate the fractions:
   \[
   59(s-2)r = 58s(r-2)
   \]

7. Distribute and simplify:
   \[
   59sr - 118r = 58sr - 116s
   \]

8. Move all terms involving $r$ to one side and all terms involving $s$ to the other side:
   \[
   59sr - 58sr = 118r - 116s
   \]

9. Combine like terms:
   \[
   sr = 118r - 116s
   \]

10. Solve for $r$:
    \[
    sr - 118r = -116s
    \]
    \[
    r(s - 118) = -116s
    \]
    \[
    r = \frac{-116s}{s - 118}
    \]

11. Since $r$ must be a positive integer, the denominator $s - 118$ must be negative (because $-116s$ is negative), which implies:
    \[
    s < 118
    \]

12. To find the largest possible value of $s$, we test the largest integer less than 118, which is 117:
    \[
    s = 117
    \]

13. Substitute $s = 117$ back into the equation for $r$:
    \[
    r = \frac{-116 \cdot 117}{117 - 118} = \frac{-13572}{-1} = 13572
    \]

14. Since $r = 13572$ is a positive integer, $s = 117$ is valid.

The final answer is $\boxed{117}$