$19.1 .15 \star \star$ Find all such natural numbers that are equal to the square of the number of their positive divisors.
Let $n=m^{2}$ have $m$ distinct divisors. If $m=1$, then $n=1$, and 1 satisfies the condition. If $m>1$, then the divisors of $n$ different from $m$ can be paired, with the product of each pair equal to $n$. Let the number of such pairs be $k$, then $m=2 k+1$ is odd, and $n=(2 k+1)^{2}$ has $k$ divisors less than $m$, all of which are odd, so $2 k-1$ is a divisor of $n=(2 k+1)^{2}$. But consecutive odd numbers $2 k-1, 2 k+1$ are coprime, so it must be that $2 k-1=1$, i.e., $k=1$. In this case, $n=(2 k+1)^{2}=9$.
Therefore, the natural numbers that satisfy the condition are only 1 and 9.