Please check for my idea : If $k\leq 3$, then there are at most three elements in $A$, let them be $a<b<c$ (because I am too lasy to type $a_{1},a_{2},a_{3}...$ in LaTex ), and $A$ can only "produce" not more than $12$ integer range between $1$ and $19$, i.e. $a,b,c,2a,2b,2c,a+b,a+c,b+c,c-a,c-b,b-a$, so $k\geq 4$. For $k=4$,let the $4$ elements be $a<b<c<d$, then there are at most $20$ integers then range between $1$ and $19$ which is "produce" from $A$,i.e. $X=\{a,b,c,d,2a,2b,2c,2d,a+b,a+c,a+d,b+c,b+d,c+d,d-a,d-b,d-c,c-b,c-a,b-a\}$ so there are at most one pair of overlap elements xor one out-of-range element in $X$. As $2d$ should not less than $19$(or else no other combination is big enough to produce $19$) but $2d\neq 19$,we know no two element in $X$ can overlap and $d+c=19$. On the other hand,$c\geq 3$ so $10\leq d \leq16$ and $c=19-d$. Case 1:$d=10,c=9$,we have $b=7$ ,try all possibe $a$,no solution. Case 2:$d=11,12,13,14,15,16$ and $c=19-d$,only $d+b$ can produce $18$,so $b=c-1$, try all possible $a$ ,no solution. For $k=5$,$A=\{2,6,7,8,11\}$ works. P.S.1 My solution to prove $k\neq 4$ is extremely ugly,hope someone can improve it. P.S.2 Consider all possibe parity combination for $k=4$ and with the fact that there are 10 odd number in $M$,we can show there should be 2 odd and 2 even in $A$ so this can reduce the times of try and error.

ychjae wrote: Please check for my idea : If $k\leq 3$, then there are at most three elements in $A$, let them be $a<b<c$ (because I am too lasy to type $a_{1},a_{2},a_{3}...$ in LaTex ), and $A$ can only "produce" not more than $12$ integer range between $1$ and $19$, i.e. $a,b,c,2a,2b,2c,a+b,a+c,b+c,c-a,c-b,b-a$, so $k\geq 4$. For $k=4$,let the $4$ elements be $a<b<c<d$, then there are at most $20$ integers then range between $1$ and $19$ which is "produce" from $A$,i.e. $X=\{a,b,c,d,2a,2b,2c,2d,a+b,a+c,a+d,b+c,b+d,c+d,d-a,d-b,d-c,c-b,c-a,b-a\}$ so there are at most one pair of overlap elements xor one out-of-range element in $X$. As $2d$ should not less than $19$(or else no other combination is big enough to produce $19$) but $2d\neq 19$,we know no two element in $X$ can overlap and $d+c=19$. On the other hand,$c\geq 3$ so $10\leq d \leq16$ and $c=19-d$. Case 1:$d=10,c=9$,we have $b=7$ ,try all possibe $a$,no solution. Case 2:$d=11,12,13,14,15,16$ and $c=19-d$,only $d+b$ can produce $18$,so $b=c-1$, try all possible $a$ ,no solution. For $k=5$,$A=\{2,6,7,8,11\}$ works. P.S.1 My solution to prove $k\neq 4$ is extremely ugly,hope someone can improve it. P.S.2 Consider all possibe parity combination for $k=4$ and with the fact that there are 10 odd number in $M$,we can show there should be 2 odd and 2 even in $A$ so this can reduce the times of try and error. After getting $d+c=19$ you can directly say $c=9$ because second biggest element of $X$ must be $2c$ or $b+d$ but $17\neq 2c$ so $2c=18 \implies c=9$. After that we can easily get $b=7$ and $a=4$ which gives a contradiction ($d+a=2b$). I did the same for the other parts of the problem. I honestly don't think there is an easier way. Another example for $k=5$: $A=\{1,2,4,10,17\}$