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Let's consider the possible ranges for each digit.
Since the maximum sum of two digits is $17$, the hundreds digit must be at most $1$ and the thousands digit must be at most $9$ (since there must be carrying).
Since the hundreds digit is at most $1$, it must be $0$.
Since the thousands digit is at most $9$, it must be at least $5$ (because there is carrying).
The remaining four digits are at most $4$, so the maximum sum of these four digits is $4+4+4+4=16$.
The smallest possible sum is $5+16=\boxed{21}$.