So sorry for my spam post, but I'm really interested in this problem. I hope a soltuion for it.

Consider $ n=10^{400}+1$, then $ \sqrt{n+k}=10^{200}\sqrt{1+\frac{k+1}{10^{400}}}$. Note that \[ 1+\frac{k}{2\cdot 10^{400}}<\sqrt{1+\frac{k+1}{10^{400}}}<1+\frac{k+1}{2\cdot 10^{400}},\] where $ 0\leq k\leq 999$. It follows that 200th digit in $ \sqrt{n+k}$ equals last digit of $ [k/2]$.

orl wrote: Is there a positive integer $ n$ such that among 200th digits after decimal point in the decimal representations of $ \sqrt {n}$, $ \sqrt {n + 1}$, $ \sqrt {n + 2}$, $ \ldots,$ $ \sqrt {n + 999}$ every digit occurs 100 times? Proposed by A. Golovanov Yes, there is. I suggest to look for a natural number $ a$ such that : $ a + p10^{ - 200}\le\sqrt {a^2 + 100p + k} < a + (p + 1)10^{ - 200}$ $ \forall p\in[0,9]$ $ \forall k\in[0,99]$ If such "$ a$" exists, then $ n = a^2$ is an answer to the required problem. First part is $ a + p10^{ - 200}\le\sqrt {a^2 + 100p + k}$ $ \iff$ $ 2ap10^{ - 200} + p^210^{ - 400}\le 100p + k$ $ \iff$ $ a\le 50\cdot 10^{200} + \frac {k10^{200}}{2p} - \frac 12p10^{ - 200}$ and so, since this must be true $ \forall p\in[0,9]$ $ \forall k\in[0,99]$, it would be enough to have : $ a\le 50\cdot 10^{200} - \frac 9210^{ - 200}$ and so $ a < 5\cdot 10^{201}$ Second part is $ \sqrt {a^2 + 100p + k} < a + (p + 1)10^{ - 200}$ $ \iff$ $ 100p + k < 2a(p + 1)10^{ - 200} + (p + 1)^210^{ - 400}$ $ \iff$ $ 50\cdot10^{200} + \frac {k - 100}{2(p + 1)}10^{200} - \frac 12(p + 1)10^{ - 200} < a$ and so, since this must be true $ \forall p\in[0,9]$ $ \forall k\in[0,99]$, it would be enough to have : $ a > 50\cdot10^{200} - \frac 1210^{199} - \frac 1210^{ - 200}$ And so $ a\ge 50\cdot 10^{200} - \frac 1210^{199}$ $ = 4995\cdot 10^{198}$ So we got $ 4995\cdot 10^{198}\le a < 5000\cdot 10^{198}$ And so, for example, $ \boxed{n = 4999^210^{396}}$ would be an answer.