Let $a$, $b$, $c$ be positive integers such that $29a + 30b + 31c = 366$. Find $19a + 20b + 21c$.
1. We start with the given equation:
   \[
   29a + 30b + 31c = 366
   \]
   This equation represents the total number of days in a leap year, where \(a\), \(b\), and \(c\) are the number of months with 29, 30, and 31 days respectively.

2. In a leap year, there are 12 months. We know that:
   - February has 29 days (since it is a leap year).
   - Four months have 30 days (April, June, September, November).
   - Seven months have 31 days (January, March, May, July, August, October, December).

3. Therefore, we can assign:
   \[
   a = 1, \quad b = 4, \quad c = 7
   \]

4. We need to find the value of \(19a + 20b + 21c\). Substituting the values of \(a\), \(b\), and \(c\):
   \[
   19a + 20b + 21c = 19(1) + 20(4) + 21(7)
   \]

5. Calculate each term separately:
   \[
   19 \cdot 1 = 19
   \]
   \[
   20 \cdot 4 = 80
   \]
   \[
   21 \cdot 7 = 147
   \]

6. Add these results together:
   \[
   19 + 80 + 147 = 246
   \]

Conclusion:
\[
\boxed{246}
\]