Answer is $n=101^k$ for $k\ge 1$. Clearly if $n$ is even the condition do not hold: take $x=n+1$ and observe that $n+102$ is not coprime to $n$. This leaves us the case $n$ is odd. Now assume $101\nmid n$. Set $x\equiv -101\pmod{n}$. If $(x,n)=d$ then $d\mid 101$. Since $101$ is a prime not dividing $n$, this forces $(x,n)=1$. However, $x+101$ is divisible by $n$. Finally, we are left with the case $n$ is divisible by $101$. Now if $n$ has a prime divisor $p$ other than $101$ with $p^j\mid \mid n$ then take $x\equiv -101\pmod{p^j}$ and $x\equiv 1\pmod{n/p^j}$. Such an $x$ clearly exists by the CRT and for this $x$, $x+101$ is divisible by $p^j$, hence not coprime to $n$. Finally, any power of $101$ works, concluding the proof.

For a prime number q which divides n.(q is not equal to 101.) x is not 0 in mod q. If x is -101 in mod q. We will get a contradiction from the second condition. For the all prime numbers q except 101, there are infinitly many numbers x which is not 0 in mod q. But it is -101 in mod q. So n's only prime divisor can be 101. It satisfies the condition because if n is in form of 101^a, x is not divisible 101, so x+101 is not either. So we get n=101^a

Note that $101$ is a prime. If $n$ is not a power of $101$,consider the smallest prime $p$ apart from $101$ that divides $n$. Now all number less than $p$ are coprime to $n$. Now by our choice of $p$,we can find a number $x$ less than $p$ such that $p|101+x$,which is a contradiction. Now let $n=101^a$.We see that $n$ clearly works.

Let $p\ne 201$ be a prime divisior of $n$.Take any integer $x$ such that $\gcd(x,n)=1$.Then for any such integer $\gcd(x,p)=1$. Given condition says that, $$\gcd(x+101,p)=1\implies (x+2\times 101,p)=1\implies \gcd(x+3\times 101)=1\implies\dots$$Inductively, $\gcd(x+n\times101,p)=1\quad\forall n$. But as $\gcd(101,p)=1$ so $\{x+n\times 101\}_{n=0}^{p-1}$ forms a complete residue class $\pmod{p}$.So atleast one of $x+n\times 101$ is divisible by $p$.Contradiction. So the only prime divisior of $n$ is 101.So $n=101^k$ is the only solution, which obviously satisfies the given condition.$\blacksquare$

Trivial. gcd($n,n-101$)=$1$ or $101$ (note that $101$ is a prime) .Also note that if $n \le101$ then we can take a suitable power of n so that it is greater than $101$. if the gcd=$1$ then,By the problem $(n,n-101+101=n)=1$,which is clearly a contradiction. therefore $101|n$ now let $n=101^{k}\cdot x$ where $x$ is relatively prime to $101$. we will prove that $x=1$ for contradiction assume $x>1$ Now consider the numbers $n,x-101$ (once again if $x\le101$ then you can take a suitable power of $x$).If there's a prime $p$ dividing the two then $p|n$ $p|101^k\cdot(x-101)=101^k x-101^{k+1}=n-101^{k+1}$. therefore $p=101$ but $p|x-101$ then $p=101|x$ which is a contradiction. so gcd$(n,x-101)$=$1$. implying gcd $(n,x-101+101=x)=1$,which is once again a contradiction,proving that $x=1$. giving,$n=101^k$ and it is easily seen that all such $n$ work.

If $101\nmid n$ then obviously $\gcd(101(n-1),n)=1$ so $\gcd(101n,n)=1$ which is clealy a contradiction. If there exist $p\neq 101$ dividing $n$ then by dirichlet's, there exists prime number $x>n$ such that $x\equiv -101\pmod{p}$. Clearly, $\gcd(x,n)=1$ which implies $\gcd(x+101,n)=1$. But $p$ divides both $x+101$ and $n$, a contradiction. So $n=101^k$ where $k$ is any positive integer, which clearly works since if $\gcd(x,n)=1$ then $101\nmid x$ then $101\nmid x+101$ as well. This means that $\gcd(x+101,n)=1$, and we're done.