A positive integer $N$ has $20$ digits when written in base $9$ and $13$ digits when written in base $27$. How many digits does $N$ have when written in base $3$?

[i]Proposed by Aaron Lin[/i]
1. We start by interpreting the given conditions for the number \( N \). The first condition states that \( N \) has 20 digits when written in base 9. This implies:
   \[
   9^{19} \leq N < 9^{20}
   \]
   This is because a number with \( d \) digits in base \( b \) satisfies \( b^{d-1} \leq N < b^d \).

2. The second condition states that \( N \) has 13 digits when written in base 27. This implies:
   \[
   27^{12} \leq N < 27^{13}
   \]

3. Next, we convert these inequalities to base 3. Since \( 9 = 3^2 \) and \( 27 = 3^3 \), we can rewrite the inequalities as follows:
   \[
   9^{19} = (3^2)^{19} = 3^{38}
   \]
   \[
   9^{20} = (3^2)^{20} = 3^{40}
   \]
   Therefore, the first condition becomes:
   \[
   3^{38} \leq N < 3^{40}
   \]

4. Similarly, for the second condition:
   \[
   27^{12} = (3^3)^{12} = 3^{36}
   \]
   \[
   27^{13} = (3^3)^{13} = 3^{39}
   \]
   Therefore, the second condition becomes:
   \[
   3^{36} \leq N < 3^{39}
   \]

5. Combining these two sets of inequalities, we get:
   \[
   3^{38} \leq N < 3^{39}
   \]

6. A number \( N \) that satisfies \( 3^{38} \leq N < 3^{39} \) has 39 digits when written in base 3. This is because a number with \( d \) digits in base \( b \) satisfies \( b^{d-1} \leq N < b^d \).

Conclusion:
\[
\boxed{39}
\]