jmc

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

For $5\le n\le 9$, $\lceil \sqrt{n}\rceil =3$. There are $5$ values of $3$ in this range.

For $10\le n\le 16$, $\lceil \sqrt{n}\rceil =4$. There are $7$ values of $4$ in this range.

For $17\le n\le 25$, $\lceil \sqrt{n}\rceil =5$. There are $9$ values of $5$ in this range.

For $26\le n\le 29$, $\lceil \sqrt{n}\rceil =6$. There are $4$ values of $6$ in this range.

Consequently, our total sum is $5\cdot 3+7\cdot 4+9\cdot 5+4\cdot 6=\boxed{112}$.