Find the minimal positive integer $m$, so that there exist positive integers $n>k>1$, which satisfy
$11...1=11...1.m$, where the first number has $n$ digits $1$, and the second has $k$ digits $1$.
1. We start with the given equation \( 11\ldots1 = 11\ldots1 \cdot m \), where the first number has \( n \) digits of 1, and the second has \( k \) digits of 1. This can be written as:
   \[
   \frac{10^n - 1}{9} = \frac{10^k - 1}{9} \cdot m
   \]
   Simplifying, we get:
   \[
   m = \frac{10^n - 1}{10^k - 1}
   \]

2. For \( m \) to be an integer, \( 10^k - 1 \) must divide \( 10^n - 1 \). This implies:
   \[
   10^k - 1 \mid 10^n - 1
   \]

3. Using the property of divisibility, we know that if \( 10^k - 1 \mid 10^n - 1 \), then:
   \[
   10^k - 1 \mid 10^n - 1 - 10^{n-k}(10^k - 1) = 10^{n-k} - 1
   \]
   Repeating this process, we reduce the exponent by \( k \) each time until we get:
   \[
   10^k - 1 \mid 10^A - 1
   \]
   where \( A = n \mod k \).

4. For \( 10^k - 1 \) to divide \( 10^A - 1 \), \( A \) must be 0 (since \( 0 \leq A < k \)). Therefore, \( n \) must be a multiple of \( k \), i.e., \( n = Bk \) for some integer \( B \).

5. Substituting \( n = Bk \) into the expression for \( m \), we get:
   \[
   m = \frac{10^{Bk} - 1}{10^k - 1}
   \]
   Using the formula for the sum of a geometric series, we can write:
   \[
   m = \sum_{i=0}^{B-1} (10^k)^i
   \]

6. To minimize \( m \), we need to find the smallest \( B \) and \( k \) such that \( n > k > 1 \). The smallest \( B \) we can take is 2 (since \( B = 1 \) would make \( m = 1 \), which is not allowed as \( n > k > 1 \)).

7. The smallest \( k \) we can take is 2. Therefore, we have:
   \[
   m = \sum_{i=0}^{1} (10^2)^i = 1 + 10^2 = 1 + 100 = 101
   \]

8. Verifying, we see that:
   \[
   1111 = 11 \cdot 101
   \]
   which confirms that \( m = 101 \) is correct.

The final answer is \( \boxed{101} \)