Solution: Denote $D(k)$ as the number of digits of a given integer, $k$. Now we can approach within a simple bounding technique. If $D(a)=n$, then \[\ 2n-1 \le D(a^2) \le 2n \] and \[\ 3n-2 \le D(a^3) \le 3n+3 \] By simply considering their lower and upper bounds, that is, considering the square and cube of: $10^{x-1}$ and $10^{x+1}$. Since $n$ must be an integer, we must have $D(a^3)=3n+1$. Implying that $4n+1=2001 \Rightarrow n=500$. By writing all the $500-$digit numbers in scientific form, we have: \[\ D(a)=a_1.a_2....a_{499}*10^{499}=500\] This implies \[\ D(a^3)= D((a_1.a_2....a_{499})^3*10^{1497})=D(a_1.a_2....a_{499})+1497\] Where $0< a_1.a_2...a_{499} <10 $ (The little dot that cannot be viewed properly is not "multiply") We therefore must have $D((a_1.a_2....a_{499})^3)=4$, but this is impossible since $D(10^3)=4$. Hence, \[\ (D((a_1.a_2....a_{499})^3)+1497)+(D(a)) \le 1500+500 \le 2000 < 2001 \] Thus, one cannot find a number $a$ such that, $n+m$ equal to $2001$. $\Box$

Let $X$ be a natural number and let $X=a\cdot10^{n}$ where $a$ is rational number : $1\le a<10$ and $n+1$ is the number of digits of $X$. Than $X^3=a^3\cdot10^{3n}$. From $10^{3n}$ we get $3n+1$ digits. From $a^3$ we can get no digits (if $a<\sqrt[3]{10}$), one (if $\sqrt[3]{10}\le a<\sqrt[3]{100}$) or two (if $\sqrt[3]{100}\le a$). So the sum of number of digits of $X$ and $X^3$ is $n+1+3n+1=4n+2$, $n+1+3n+1+1=4n+3$ or $n+1+3n+1+2=4n+4$. But $2001$ couldn't be represented in any of these form. So the answer is NO. Actually $2001=4\cdot500+1$. If we replace $2001$ with $2011$ we can take $X=4\cdot 10^{502}$. We obtain that sum of number of digits of $4\cdot 10^{502}$ and $\left(4\cdot 10^{502}\right)^3$ is equal to $2011$.