10. From $1,2,3, \cdots, 20$ choose $k$ different numbers to form a set $M$, such that the set $M$ must contain 4 different numbers $a, b, c, d$ satisfying $a-b+c-d$ is divisible by 20, then the minimum value of $k$ is $\qquad$.
In $1,2,3, \cdots, 20$, select 7 numbers. The sum of these 7 numbers in pairs results in 21 numbers. Since there are 20 numbers for the remainder modulo 20, there exist two pairs of sums with the same remainder modulo 20.
If these two pairs of sums are $a+c \equiv b+d(\bmod 20)$ (where $a, b, c, d$ are 4 different numbers), then $20 \mid(a-b+c-d)$, satisfying the condition;
If these two pairs of sums are $a+x \equiv b+x(\bmod 20)$, then $a \equiv b(\bmod 20)$, which contradicts $a, b \in\{1,2,3, \cdots, 20\}$. On the other hand, take $M=\{1,2,4,7,12,20\}$, then the set of pairwise sums of numbers in $M$ is $\{3,5,8,13,1,6,9,14,2,11, 16,4,19,7,12\}$, clearly there are no 4 numbers that satisfy the condition.
Therefore, the minimum value of $k$ is 7.