jmc

Since the expression $\lceil \sqrt{x}\rceil$ stands for the least integer that is greater than or equal to $x$, the largest possible value of $x$ that could satisfy the equation is $15^2$, or $225$. The greatest integer smaller than $15$ is $14$, so the largest integer (smaller than $225$) that would not satisfy $\lceil \sqrt{x}\rceil =15$ would be $14^2$, or $196$. Therefore, any integer that is in the range $196<x\le 225$ could be considered a possible integer value of $x$. Since there are 29 numbers in this range, our final solution is $\boxed{29}$.