2 Find all positive integers $n$, such that the decimal integer composed of $n-1$ digits 1 and one digit 7 is a prime number.

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2. A positive integer $N$ composed of $n-1$ digits 1 and one digit 7 can be expressed in the form $N = A_{n} + 6 \times 10^{k}$, where $0 \leqslant k \leqslant n-1$, and $A_{n}$ is an integer composed of $n$ ones.

When $3 \mid n$, the sum of the digits of $A_{n}$ is divisible by 3, so $3 \mid A_{n}$, and thus $3 \mid N$. Since $N > 3$, $N$ is not a prime number in this case.

Now, let's assume $3 \nmid n$. Note that $10^{6} \equiv 1(\bmod 7)$. We will classify $n$ modulo 6 to discuss the value of $A_{n}$ modulo 7 (the cases $n \equiv 0,3(\bmod 6)$ do not need to be considered). It is easy to see that for $l \geqslant 0$,
$$\begin{aligned}
A_{6l+1} & =\frac{1}{9} \times\left(10^{6l+1}-1\right)=\frac{1}{9} \times\left(10^{6l}-1\right) \times 10+\frac{1}{9} \times(10-1) \\
& \equiv 1(\bmod 7) \\
A_{6l+2} & \equiv 4, \quad A_{6l+4} \equiv 5, \quad A_{6l+5} \equiv 2(\bmod 7)
\end{aligned}$$

Additionally, $10^{0}, 10^{2}, 10^{4}, 10^{5}$ are congruent to $1,2,4,5$ modulo 7, respectively. Therefore, when $n > 6$, for $n \equiv 1,2,4,5(\bmod 6)$, we can take $k=0,4,5,2$ respectively, and we find that
$$N = A_{n} + 6 \times 10^{k} \equiv A_{n} - 10^{k} \equiv 0(\bmod 7),$$

hence $N$ is not a prime number, and thus $n > 5$ does not meet the requirement. When $n \leqslant 5$, it is not difficult to verify that only $n = 1,2$ meet the requirement.