Each term of a sequence of natural numbers is obtained from the previous term by adding to it its largest digit. What is the maximal number of successive odd terms in such a sequence?
To determine the maximal number of successive odd terms in a sequence where each term is obtained by adding its largest digit to the previous term, we need to analyze the behavior of the sequence based on the last digit of each term.

1. **Initial Considerations**:
   - If a term ends in 9, the next term will be even because adding 9 to any number results in a number ending in 8.
   - Therefore, a term ending in 9 cannot be followed by an odd term.

2. **Case Analysis**:
   We will consider the cases where the first term \( t_1 \) ends in 1, 3, 5, 7, or 9.

   **Case 1: \( t_1 \) ends in 1**:
   - The largest digit \( l \) of \( t_1 \) must be even (since \( t_2 \) must be odd).
   - Adding an even digit to a number ending in 1 results in a number ending in an odd digit.
   - However, the next term \( t_3 \) will end in an even digit because adding the largest digit (which is odd) to an odd number results in an even number.
   - Therefore, at most 2 successive odd terms can exist.

   **Case 2: \( t_1 \) ends in 3**:
   - If the largest digit \( l \) of \( t_1 \) is such that \( l + 3 \leq 9 \), then \( t_2 \) will end in \( l + 3 \), which is even.
   - If \( l + 3 > 9 \), then \( l = 8 \) and \( t_2 \) will end in 1.
   - Using Case 1, at most 2 more terms can follow, resulting in at most 3 successive odd terms.

   **Case 3: \( t_1 \) ends in 5**:
   - If the largest digit \( l \) of \( t_1 \) is such that \( l + 5 \leq 9 \), then \( t_2 \) will end in \( l + 5 \), which is even.
   - If \( l + 5 > 9 \), then \( l = 6 \) and \( t_2 \) will end in 1.
   - Using Case 1, at most 2 more terms can follow, resulting in at most 3 successive odd terms.
   - If \( l = 8 \), then \( t_2 \) will end in 3.
   - Using Case 2, at most 2 more terms can follow, resulting in at most 4 successive odd terms.

   **Case 4: \( t_1 \) ends in 7**:
   - If the largest digit \( l \) of \( t_1 \) is such that \( l + 7 \leq 9 \), then \( t_2 \) will end in \( l + 7 \), which is even.
   - If \( l + 7 > 9 \), then \( l = 4 \) and \( t_2 \) will end in 1.
   - Using Case 1, at most 2 more terms can follow, resulting in at most 3 successive odd terms.
   - If \( l = 6 \), then \( t_2 \) will end in 3.
   - Using Case 2, at most 2 more terms can follow, resulting in at most 4 successive odd terms.
   - If \( l = 8 \), then \( t_2 \) will end in 5.
   - Using Case 3, at most 3 more terms can follow, resulting in at most 5 successive odd terms.

   **Case 5: \( t_1 \) ends in 9**:
   - The next term will be even, as adding 9 results in a number ending in 8.
   - Therefore, no successive odd terms can follow.

3. **Conclusion**:
   - The maximum number of successive odd terms in the sequence is 5.
   - An example sequence is \( 817, 825, 833, 841, 849 \).

The final answer is \(\boxed{5}\).