Let A = ∑ i = 1 10 ∑ j = 1 10 min i , j and B = ∑ i = 1 10 ∑ j = 1 10 max i , j .…

$A=\sum _{i=1}^{10}\sum _{j=1}^{10}\min \left\{i,j\right\}$

$B=\sum _{i=1}^{10}\sum _{j=1}^{10}\max \left\{i,j\right\}$

$A=\sum _{j=1}^{10}\min \left(i,1\right)+\min \left(j,2\right)+\dots \min \left(i,10\right)$

$=\underset{19 \mathrm{times} }{\underbrace{\left(1+1+1+\dots +1\right)}}+\underset{17 \mathrm{times} }{\underbrace{\left(2+2+2\dots +2\right)}}+\underset{15 \mathrm{times} }{\underbrace{\left(3+3+3\dots +3\right)}} +...$

$B=\sum _{j=1}^{10}\max \left(i,1\right)+\max \left(j,2\right)+\dots \max \left(i,10\right)$

$=\underset{19 \mathrm{times} }{\underbrace{\left(10+10+\dots +10\right)}}+\underset{17 \mathrm{times} }{\underbrace{\left(9+9+\dots +9\right)}}+...$

i.e. each element of the ordered pair $\left\{i,j\right\}$ will appear either in $A \text{or} B$

So $A+B$ will be sum of all elements of the ordered pair 

$A+B=20\left(1+2+3+\dots +10\right)$

$=20\times \frac{10\times 11}{2}=10\times 110=1100$