Inspired by Brian Charlesworth

$2x+y=10$

$2(x+y)=10+y$

$(x+y)=5+ \dfrac{y}{2}$

So, $(x+y)_{max}$ is possible when $y$ is maximised

$\because y_{max}=10$

$\therefore (x+y)_{max}=5+ \dfrac{10}{2}=10$

$(x+y)_{min}$ is possible when $y=0$

$(x+y)_{min}= 5$

$(x+y)_{max}+(x+y)_{min} = 10+5 = \boxed{15}$

The previously mentioned answers are impressive (for me at least) and much appreciated, but for a simpler approach (and for a simpler mind such as mine):

Taking into consideration that 'x' and 'y' are non-negative integers, therefore:

The minimum value for 'x' is 0, so solving the equation: 2(0) + y = 10 —> y=10

The minimum value for 'y' is 0, so solving the equation: 2x + (0) = 10 —> x=5

Hence, the sum of minimum and maximum values of (x+y)= [(0+10)+(5+0)]=15

Note that $2x+y=10$ is a Diophantine equation that has integer solutions. Hence, the solutions for $x$ and $y$ are $x=x_{0}-r, y=y_{0}+2r$ for some integer $r$ . Hence the minimum value of $x+y$ is 5 and the maximum is 10, so we get the answer 15.

2X+Y=10

First non negative integer is 0. Put it for X Y=10

Put Y=0 2X=10 X=5 Therefore X+Y=5+10=15

One may try Linear (or Integer) programming... Just suggesting.