Denote the equation in the problem statement as (*), and note that it is equivalent to the condition that the average of the remainders when dividing $n, 2 n, \ldots, k n$ by $202$ is $101$. Since $\left\{\frac{i n}{202}\right\}$ is invariant in each residue class modulo $202$ for each $1 \leq i \leq k$, it suffices to consider $0 \leq n<202$.

If $n=0$, so is $\left\{\frac{i n}{202}\right\}$, meaning that $(*)$ does not hold for any $k$. If $n=101$, then it can be checked that $(*)$ is satisfied if and only if $k=1$. From now on, we will assume that $101 \nmid n$.

For each $1 \leq i \leq k$, let $a_{i}=\left\lfloor\frac{i n}{202}\right\rfloor=\frac{i n}{202}-\left\{\frac{i n}{202}\right\}$. Rewriting $(*)$ and multiplying the equation by $202$, we find that
$$n(1+2+\ldots+k)-202\left(a_{1}+a_{2}+\ldots+a_{k}\right)=101 k$$
Equivalently, letting $z=a_{1}+a_{2}+\ldots+a_{k}$,
$$n k(k+1)-404 z=202 k$$
Since $n$ is not divisible by $101$, which is prime, it follows that $101 \mid k(k+1)$. In particular, $101 \mid k$ or $101 \mid k+1$. This means that $k \in\{100,101,201\}$. We claim that all these values of $k$ work.

- If $k=201$, we may choose $n=1$. The remainders when dividing $n, 2 n, \ldots, k n$ by $202$ are $1,2, \ldots, 201$, which have an average of $101$.

- If $k=100$, we may choose $n=2$. The remainders when dividing $n, 2 n, \ldots, k n$ by $202$ are $2,4, \ldots, 200$, which have an average of $101$.

- If $k=101$, we may choose $n=51$. To see this, note that the first four remainders are $51,102,153,2$, which have an average of $77$. The next four remainders ($53,104,155,4$) are shifted upwards from the first four remainders by $2$ each, and so on, until the $25$th set of the remainders ($99,150,201,50$) which have an average of $125$. Hence, the first $100$ remainders have an average of $\frac{77+125}{2}=101$. The $101$st remainder is also $101$, meaning that the average of all $101$ remainders is $101$.

In conclusion, all values $k \in\{1,100,101,201\}$ satisfy the initial condition.