After recalculating the table I believe that there is a winning strategy for A, as follows:
If D(i) is the number of divisors of i, and C(i) is the relative benefit of moving to i then C(i) = 0 for i >= 100 and C(i) = D(i) - Max(C(i+1), C(i+2),...C(i+10)) for 1 <= i <= 99. You can use this to move backwards down the list of numbers from 99 to 1, calculating C(i) at each step. Because A moves first they have a choice if 'starting' at any i between 1 and 10. The table of C(i) below confirms that A can win by moving to 1.
A chooses 1 first and ends up winning by 1 point.
See Tables below showing the way an optimal game could progress. Spoiler space added as I cant work out how to hide a table.
An optimal game would progress as follows (Asc and Bsc are the scores for A and B after each move):
Mov Pl. Num Sum Asc BSc
1 A 1 1 1 0
2 B 9 10 1 4
3 A 2 12 7 4
4 B 1 13 7 6
5 A 7 20 13 6
6 B 4 24 13 14
7 A 6 30 21 14
8 B 6 36 21 23
9 A 6 42 29 23
10 B 7 49 29 26
11 A 7 56 37 26
12 B 4 60 37 38
13 A 1 61 39 38
14 B 9 70 39 46
15 A 2 72 51 46
16 B 8 80 51 56
17 A 10 90 63 56
18 B 9 99 63 62
The above was calculated using the following table showing the relative advantage in moving to any specific position (calculated using the algorithm described above).
Sum Div Adv
1 1 1
2 2 -2
3 2 -2
4 3 -1
5 2 -2
6 4 0
7 2 -2
8 4 0
9 3 -1
10 4 0
11 2 -2
12 6 4
13 2 2
14 4 -2
15 4 -2
16 5 -1
17 2 -4
18 6 0
19 2 -4
20 6 0
21 4 -2
22 4 -2
23 2 -4
24 8 6
25 3 1
26 4 -2
27 4 -2
28 6 0
29 2 -4
30 8 2
31 2 -4
32 6 0
33 4 -2
34 4 -2
35 4 -2
36 9 6
37 2 -1
38 4 -1
39 4 -1
40 8 3
41 2 -3
42 8 3
43 2 -3
44 6 1
45 6 1
46 4 -1
47 2 -3
48 10 5
49 3 5
50 6 -4
51 4 -6
52 6 -4
53 2 -8
54 8 -2
55 4 -6
56 8 -2
57 4 -6
58 4 -6
59 2 -8
60 12 10
61 2 2
62 4 -4
63 6 -2
64 7 -1
65 4 -4
66 8 0
67 2 -6
68 6 -2
69 4 -4
70 8 0
71 2 -6
72 12 8
73 2 -2
74 4 -2
75 6 0
76 6 0
77 4 -2
78 8 2
79 2 -4
80 10 4
81 5 -1
82 4 -2
83 2 -4
84 12 6
85 4 -2
86 4 -2
87 4 -2
88 8 2
89 2 -4
90 12 6
91 4 -2
92 6 0
93 4 -2
94 4 -2
95 4 -2
96 12 6
97 2 -4
98 6 0
99 6 6