jmc

Since $1\le \sqrt{1}<\sqrt{2}<\sqrt{3}<2,$ the first three terms of the sum are equal to $1.$ Then, since $2\le \sqrt{4}<\sqrt{5}<\cdots <\sqrt{8}<3,$ the next five terms equal $2.$ Then, since $3\le \sqrt{9}<\sqrt{10}<\cdots <\sqrt{15}<4,$ the next seven terms equal $3.$ Finally, the last term equals $\lfloor 4\rfloor =4.$ So the overall sum is $$3(1)+5(2)+7(3)+4=3+10+21+4=\boxed{38}.$$