10. (10 points) Find the largest positive integer that is divisible by 7 and has all odd digits, with the sum of its digits being 2017.
【Analysis】To make the integer as large as possible, and each digit is odd, it must ensure that the number of digits of the integer is sufficient, and contains as many 1s as possible; according to this analysis, the solution can be derived.

【Solution】To make the integer as large as possible, and each digit is odd, it must ensure that the number of digits of the integer is sufficient, and contains as many 1s as possible.

According to the characteristics of numbers divisible by 7, 111111 is the smallest number with each digit being 1 and divisible by 7.
Also: $2017=6 \times 336+1=6 \times 335+7$
When there are 336 111111s, because the sum of all digits must be 2017, the first digit can only be 1, which cannot be divisible by 7;

When there are 335 111111s, a positive integer needs to be added at the front, so that the sum of its digits equals 7, and this number is as large as possible. The largest integer that meets this condition is 13111.

Explanation: We can use the following method to construct a sequence of numbers that can be divided by 7 and, except for the first digit, the rest are 1:
21, $490+21=511$, $700+511=1211$, $5600+511=6111$, $7000+6111=13111$, $35000+6111=41111$, $70000+41111=111111$, $70000+41111=111111$,
We notice that $7000+6111=13111$ is the largest positive integer that can be divided by 7 and the sum of its digits equals 7. Therefore, the largest positive integer with the sum of its digits being 2017 is $13_{2013 \text{ ones}}$, where the number of 1s is $335 \times 6+4=2014$.

Answer: The largest positive integer that can be divided by 7, with all digits being odd, and the sum of its digits being 2017 is $13_{2013 \text{ ones}}$.