$ S=(m+1)+\cdots+(m+n)=\frac{(m+n)(m+n+1)}{2}-\frac{m(m+1)}{2}=$ $ =\frac{m^2+n^2+2mn+m+n-m^2-m}{2}=\frac{n^2+2mn+n}{2}=$ $ =\frac{n(n+2m+1)}{2}$ if $ n=2k+1$ then $ S=(2k+1)(k+m+1)$ take $ m=k$. then for every $ k$ there are $ 2k+1$ consecutive integers such as: $ (k+1)+(k+2)+\cdots+(k+2k+1)=(2k+1)^2$ Example for $ k=2$, $ n=5$ $ 3+4+5+6+7=5^2$