Let $a$ and $b$ be positive integers satisfying
	\[ \frac a{a-2} = \frac{b+2021}{b+2008} \]
Find the maximum value $\dfrac ab$ can attain.
1. Start with the given equation:
   \[
   \frac{a}{a-2} = \frac{b+2021}{b+2008}
   \]

2. Cross-multiply to eliminate the fractions:
   \[
   a(b + 2008) = (a - 2)(b + 2021)
   \]

3. Expand both sides:
   \[
   ab + 2008a = ab + 2021a - 2b - 4042
   \]

4. Simplify by canceling out \(ab\) from both sides:
   \[
   2008a = 2021a - 2b - 4042
   \]

5. Rearrange the terms to isolate \(b\):
   \[
   2008a - 2021a = -2b - 4042
   \]
   \[
   -13a = -2b - 4042
   \]

6. Simplify further:
   \[
   13a = 2b + 4042
   \]

7. Solve for \(b\):
   \[
   2b = 13a - 4042
   \]
   \[
   b = \frac{13a - 4042}{2}
   \]

8. Since \(b\) must be an integer, \(13a - 4042\) must be even. This implies \(a\) must be even. Let \(a = 2k\) for some integer \(k\):
   \[
   b = \frac{13(2k) - 4042}{2}
   \]
   \[
   b = \frac{26k - 4042}{2}
   \]
   \[
   b = 13k - 2021
   \]

9. Since \(b\) must be a positive integer, we need:
   \[
   13k - 2021 > 0
   \]
   \[
   13k > 2021
   \]
   \[
   k > \frac{2021}{13}
   \]
   \[
   k > 155.4615
   \]

10. The smallest integer \(k\) greater than 155.4615 is 156. Thus, let \(k = 156\):
    \[
    a = 2k = 2 \times 156 = 312
    \]
    \[
    b = 13k - 2021 = 13 \times 156 - 2021 = 2028 - 2021 = 7
    \]

11. Calculate \(\frac{a}{b}\):
    \[
    \frac{a}{b} = \frac{312}{7}
    \]

12. Verify if increasing \(b\) by multiples of 13 and corresponding \(a\) by multiples of 2 results in a smaller ratio:
    \[
    b = 7 + 13n, \quad a = 312 + 2n
    \]
    \[
    \frac{a}{b} = \frac{312 + 2n}{7 + 13n}
    \]

    As \(n\) increases, the ratio \(\frac{312 + 2n}{7 + 13n}\) decreases because the numerator increases linearly while the denominator increases more rapidly.

Conclusion:
The maximum value of \(\frac{a}{b}\) is achieved when \(a = 312\) and \(b = 7\).

The final answer is \(\boxed{\frac{312}{7}}\).