15. Find the smallest positive integer $n$, such that among any $n$ consecutive positive integers, there is at least one number whose sum of digits is a multiple of 7.
15. First, we can point out 12 consecutive positive integers, for example,
$$
994,995, \cdots, 999,1000,1001, \cdots, 1005 \text {, }
$$

where the sum of the digits of any number is not a multiple of 7. Therefore, $n \geqslant 13$.

Next, we prove that in any 13 consecutive positive integers, there must be one number whose sum of digits is a multiple of 7.

For each non-negative integer $a$, we call the set of 10 numbers
$$
A_{a}=\{10 a, 10 a+1, \cdots, 10 a+9\}
$$

a "basic segment." 13 consecutive positive integers either belong to two basic segments or belong to three basic segments.

When 13 numbers belong to two basic segments, by the pigeonhole principle, there must be 7 consecutive numbers belonging to the same basic segment;

When 13 consecutive numbers belong to three basic segments $A_{a-1} 、 A_{a} 、 A_{a+1}$, there must be 10 consecutive numbers all belonging to $A_{a}$. Now, let

be 7 numbers belonging to the same basic segment, and their sums of digits are
$$
\sum_{i=0}^{k} a_{i}, \sum_{i=0}^{k} a_{i}+1, \cdots, \sum_{i=0}^{k} a_{i}+6 .
$$

Clearly, these 7 sums have different remainders when divided by 7, and one of them must be a multiple of 7.
Therefore, the smallest value is $n=13$.