11. (10 points) In a box, there are two types of beads, one weighing 19 grams each, and the other weighing 17 grams each. The total weight of all the beads is 2017 grams. Find the number of each type of bead and all possible values.


【Analysis】According to the problem, we can set the number of each weight as an unknown, with 19-gram beads having $x$ pieces and 17-gram beads having $y$ pieces, and then list the relationship, taking values step by step within the range of positive integers, and finally find the values that meet the requirements.

【Solution】Solution: According to the analysis, let there be $x$ 19-gram beads and $y$ 17-gram beads, then:
$19 x+17 y=2017$, and since $x, y$ are both positive integers,
$\therefore 1 \leqslant x \leqslant \frac{2017-17 \times 1}{19}=\frac{2000}{19}<106, \quad 1 \leqslant y \leqslant \frac{2017-19 \times 1}{17}=\frac{1996}{17}<118$;
$19 x+17 y=2017 \Rightarrow x=\frac{2017-17 y}{19}$, by the remainder theorem, to make $x$ a positive integer, $2017-17 y$ must be divisible by 19,

i.e., the remainder is 0, and 2017 divided by 19 has a remainder of 3, so $17 y$ divided by 19 also has a remainder of 3, among all numbers that have a remainder of 3 when divided by 19, are less than 2017, and can be divided by 17, there are only:
(1) 136, i.e., $17 y=136 \Rightarrow y=8, x=\frac{2017-17 \times 8}{19}=99, x+y=99+8=107$;
(2) 459, i.e., $17 y=459 \Rightarrow y=27, x=\frac{2017-459}{19}=82, x+y=82+27=109$;
(3) 782, i.e., $17 y=782 \Rightarrow y=46, x=\frac{2017-782}{19}=65, x+y=65+46=111$;
(4) 1105, i.e., $17 y=1105 \Rightarrow y=65, x=\frac{2017-1105}{19}=48, x+y=48+65=113$;
(5) 1428, i.e., $17 y=1428 \Rightarrow y=84, x=\frac{2017-1428}{19}=31, x+y=31+84=115$;
(6) 1751, i.e., $17 y=1751 \Rightarrow y=103, x=\frac{2017-1751}{19}=14, x+y=14+103=117$.

In summary, the possible values of the sum of the two types of beads, i.e., $x+y$, are: 107, 109, 111, 113, 115, 117.
Therefore, the answer is: 107, 109, 111, 113, 115, 117.