In row $i$, column $j$ you have the number $i+j+1$. You’re going to pick one number from each row. Let’s say that in row $1$ you pick the number in column $j_1$, in row $2$ you pick the number in column $j_2$, and so on. Your $n$ numbers will then be $1+j_1+1,2+j_2+1,\dots,i+j_i+1,\dots,n+j_n+1\;.$ When you add them, you’ll get
$\sum_{i=1}^n(i+j_i+1)=\sum_{i=1}^ni+\sum_{i=1}^nj_i+\sum_{i=1}^n1\;.$
You should know a formula for the first of those three summations, and the last one is trivial, so the only one that you have to worry about is the middle one,
$\sum_{i=1}^nj_i\;.$
But remember, you chose only one number from each column, so no two of the column numbers $j_i$ are the same. And you chose $n$ numbers, so you must have chosen one from every column. Thus, the numbers $j_1,j_2,\dots,j_n$ are simply the numbers $1,2,\dots,n$ in some order, and their sum is the same as the sum of the integers $1$ through $n$.
You should now be able to analyze the second problem in the same way.