Case 1: first number is $3k+1$ Then $2n$ numbers are $3k+1,3k+2,3k+4,3k+5,...,3k+3(n-1)+1,3k+3(n-1)+2$ with sum $6kn+3n^2$ $6kn+3n^2=300 \to n(2k+n)=100 \to n=2m,m<5 \to m(k+m)=25 \to m=1,k=24,n=2$ and these numbers are $73,74,76,77$ Case 2: first number is $3k+2$ Then $2n$ numbers are $3k+2,3k+4,3k+5,...,3k+3(n-1)+1,3k+3(n-1)+2,3k+3n+1$ with sum $6kn+3n^2+3n$ $6kn+3n^2+3n=300 \to n(2k+n+1)=100 \to n=4,k=10$ and numbers are $32,34,...,43$