jmc

We note that if $a^2\le n<(a+1{)}^2$ for some integer $a$, then $a\le \sqrt{x}<a+1$, so $a$ is the greatest integer less than or equal to $x$. Consequently, we break up our sum into the blocks of integers between consecutive perfect squares:

For $1\le n\le 3$, $\lfloor \sqrt{n}\rfloor =1$. There are $3$ values of $n$ in this range. For $4\le n\le 8$, $\lfloor \sqrt{n}\rfloor =2$. There are $5$ values of $n$ in this range. For $9\le n\le 15$, $\lfloor \sqrt{n}\rfloor =3$. There are $7$ values of $n$ in this range. For $16\le n\le 19$, $\lfloor \sqrt{n}\rfloor =4$. There are $4$ values of $n$ in this range.

Consequently, our total sum is $3\cdot 1+5\cdot 2+7\cdot 3+4\cdot 4=\boxed{50}$.