Let's find a natural number $n$, preferably small, such that the decimal representation of $1 / n$ has a segment of 3 consecutive identical digits, but the segment should not consist entirely of the same digit.
Preliminary remark. Solutions I-III, partially building on each other, provide progressively smaller values of $n$, but with increasingly sophisticated reasoning.

Solution I. For every $n>1000$, in $1 / n$ at least three zeros follow the decimal point, and if $n$ is not divisible by 2 or 5, then these digits, as is known, belong to the segment.

The smallest such value is $n=1001=7 \cdot 11 \cdot 13$, and indeed $1 / 1001=0.0009990009 \ldots$, satisfying the requirement with two different, three-digit sequences.

Korányi László (Budapest, Móricz Zs. Gymnasium, 3rd grade)

Solution II. We will show that the desired property cannot be met by a two-digit number, by tracking the progression of consecutive remainders in the division of 1 by $n$. Let $1 / n$ have three consecutive decimal digits $j$ somewhere in its expansion, with the remainder before the first $j$ being $r_{1}$, and the subsequent remainders being $r_{2}, r_{3}, r_{4}$, each of which is a number between 1 and $n-1$, including these limits. Then, by the trial division,

$$
10 r_{1}=j n+r_{2}
$$

where $r_{2} \neq r_{1}$, otherwise from here on every digit of $1 / n$ would be $j$, which we have excluded. Therefore,

$$
r_{2}=r_{1}+d
$$

where $d$ is an integer and $|d| \geq 1$. Furthermore, similarly,

$$
\begin{aligned}
& r_{3}=10 r_{2}-j n=10\left(r_{1}+d\right)-j n=r_{2}+10 d=r_{1}+11 d \\
& r_{4}=10 r_{3}-j n=10\left(r_{1}+11 d\right)-j n=r_{2}+110 d=r_{1}+111 d
\end{aligned}
$$

thus

$$
\left|r_{4}-r_{1}\right|=|111 d| \geq 111
$$

However, the difference between any two possible remainders between 0 and $n$ is at most $n-2$, so for $n$ to meet the problem's requirements, $n \geq 113$.

Based on this, we seek an $n$ between 100 and 1000 for which the three identical $j$ digits follow directly after two zero digits in the decimal expansion of $1 / n$:

$$
1 / n=0.00 j j j m \ldots
$$

where $m$ is a potentially different digit. This form only deviates from $0.00 j j j j \ldots$ at the sixth decimal place, with all subsequent digits also being $j$. This form, as is known, is the reciprocal of $N=900$ when $j=1$, of $N=450$ when $j=2$, and of $N=300$ when $j=3$. It is therefore natural to look for $n$ among the numbers adjacent to these $N$ values.

$n=299$ does not meet the criteria, as in $1 / 299$ the remainder before the first $j=3$ is $r_{1}=100$, followed by $r_{2}=103$, so the value of $d$ is 3, and thus $r_{4}=r_{1}+111 d=433>299$. No $n \leq 298$ meets the stricter requirement either.

For $n=301$, similarly, $d=-3$ and $r_{4}=\frac{111 j+1}{9-j}$

The possible $r_{1}$ values and the corresponding $n$ values from (4) are as follows:

For $j=1$:

$$
\begin{aligned}
r_{1} & =15, \quad 16, \quad 17, \quad 18, \ldots, 22, \\
n & =134,143,152,161, \ldots, 197
\end{aligned}
$$

For $j=2$ (4) yields only odd $r_{1}$ values that give integer $n$:

$$
\begin{aligned}
& r_{1}=33, \quad 35, \quad 37, \ldots, \quad 43 \\
& n=148, \quad 157, \quad 166, \ldots, 193
\end{aligned}
$$

Our result so far is that only the numbers from (5) and (6) can meet the problem's requirements below the chosen 200 limit, but we need to test each case to determine which numbers actually meet the criteria. A pair of $n, r_{1}$ values where $n$ is even and $r_{1}$ is odd cannot work, as dividing $10 r_{1}$ by an even $n$ results in an even remainder, which cannot return to $r_{1}$ after any number of division steps.

The smallest of the found $n$ values: 143 does not meet the criteria, as in $1 / 143$ only 6 different remainders appear: 1, 10, 100, 142, 133, 43, and $r_{1}=16$ is not among them. $n=157$ and $n=161$ do meet the criteria, with the latter being confirmed more quickly: $1 / 161=0.0062111801 \ldots$, and the 35th-37th decimal places are 2s.

Essentially with the same reasoning for $d=-1$:

$$
0111$, and from (3) by eliminating $r_{1}$

$$
\begin{aligned}
& 0\frac{1000}{n}>\frac{1000}{200}=5
\end{aligned}
$$

Thus, the repeating digit can only be $j=7$ and 8. From (4) and for $j=8$, the following pairs of $r_{1}, n$ are obtained:

$$
\begin{aligned}
& r_{1}=119,127,135,143, \ldots, 175 \\
& n=134,143,152,161, \ldots, 197
\end{aligned}
$$

For $j=7$:

$$
\begin{aligned}
& r_{1}=115,122,129,136, \ldots, 160 \\
& n=148,157,166,175, \ldots, 193
\end{aligned}
$$

We notice that we obtained the same $n$ values as in the case of $d=+1$. Indeed, by continuing the above divisions, in $1 / 161$ the 38th-40th digits are 8s and in $1 / 157$ the 74th-76th digits are 7s.

Our method also proves that no number smaller than 157 meets the problem's criteria. We also observed that in the decimal expansion of the reciprocals of 157, 161, and 1001, at least two sets of three identical digits appear consecutively.

Földes Tamás (Budapest, Berzsenyi D. Gymnasium, 3rd grade)

Remark. The interesting fact that for $n=157$ (a prime number) two sets of three identical digits appear can be explained by the following theorem. If $p$ is a prime number and the length of the segment (the number of digits in the segment) of $1 / p$ is an even number: $2 h$, then in the expansion, two digits $h$ steps apart sum to 9, because the sum of two remainders $h$ steps apart is $p$.

This property also appears for $n=1001$ and 161, although these are not primes.

Furthermore, in the expansion of $1 / 449$ (449 is also a prime), after calculating 1