No, this is not possible. To see this, let $x_i<x_j<x_k$ be such numbers with the property that, upon including the gcd of the remainder to them, we obtain the same number. Suppose, $x_j=a+x_i$ and $x_k=b+x_i$ with $b>a>0$, and let $g={\rm gcd}(x_u:u\notin\{i,j,k\})$. After adding the gcd, we get $x_i'=x_i+(g,x_i+a,x_i+b)$, and $x_k'=x_k+(g,x_i,x_i+a)=x_i+b+(g,x_i,x_i+a)$. Now, $x_i'=x_k'$ yields, $b=(g,x_i+a,x_i+b)-(g,x_i,x_i+a)$. Now, note that, $(g,x_i+a,x_i+b)\mid (x_i+a,x_i+b)$, and thus, it divides $b-a$, and hence, is less than or equal to $b-a$. This yields a contradiction.