James from Evenlode sent a very well-explained solution this problem.  He said:

To make pairs to $10$ you need to leave out $5$ and $10$, because you would need a double for the $5$ and a $0$ for the $10$.
To make pairs to $12$ you need to leave out $1$ and $6$, because you need $11$ for the $1$ and double for the $6$.
To make pairs to $13$ you need to leave out $1$ and $2$, because you need numbers higher than $10$ ($11$ and $12$) .
To make pairs to $11$ ... you don't need to leave ANYTHING out, because $11$ is an odd number so you don't need any doubles AND you don't need partners bigger than $10$.

Thank you also to Johannes, Fred and Lok from St Barnabas,  Henry from Moulsford, children from Jebel Ali Primary School, Alex from Templars Primary, Alex from Maidstone and Beth from Torpoint who all sent clear solutions.