A positive integer $k > 1$ is called nice if for any pair $(m, n)$ of positive integers satisfying the condition $kn + m | km + n$ we have $n | m$.
1. Prove that $5$ is a nice number.
2. Find all the nice numbers.
1. **Prove that \(5\) is a nice number.**

   Given: \( kn + m \mid km + n \) for \( k = 5 \), we need to show that \( n \mid m \).

   Let's start by substituting \( k = 5 \):
   \[
   5n + m \mid 5m + n
   \]

   Suppose \(\gcd(m, n) = d\), then we can write \( m = da \) and \( n = db \) where \(\gcd(a, b) = 1\). Substituting these into the divisibility condition, we get:
   \[
   5(db) + da \mid 5(da) + db
   \]
   Simplifying, we have:
   \[
   d(5b + a) \mid d(5a + b)
   \]
   Since \(d\) is a common factor, we can divide both sides by \(d\):
   \[
   5b + a \mid 5a + b
   \]

   Now, we need to show that \(5b + a \mid 5a + b\). Consider the expression:
   \[
   5a + b = (5b + a) \cdot q + r
   \]
   where \(q\) is the quotient and \(r\) is the remainder. Since \(5b + a \mid 5a + b\), the remainder \(r\) must be zero. Therefore:
   \[
   5a + b = (5b + a) \cdot q
   \]

   Rearranging, we get:
   \[
   5a + b - (5b + a) \cdot q = 0
   \]
   Simplifying further:
   \[
   5a + b - 5bq - aq = 0
   \]
   \[
   5a + b = 5bq + aq
   \]
   \[
   5a + b = q(5b + a)
   \]

   Since \(\gcd(a, b) = 1\), the only way this can hold is if \(q = 1\) and \(5b + a = 5a + b\). This implies:
   \[
   5b + a = 5a + b
   \]
   \[
   4b = 4a
   \]
   \[
   b = a
   \]

   Therefore, \(n = db\) and \(m = da\) imply \(n \mid m\).

2. **Find all the nice numbers.**

   We need to find all \(k\) such that for any pair \((m, n)\) of positive integers satisfying \(kn + m \mid km + n\), we have \(n \mid m\).

   From the previous part, we know that \(k = 5\) is a nice number. Let's generalize this for any \(k\).

   Given \(kn + m \mid km + n\), we need to show that \(n \mid m\).

   Suppose \(\gcd(m, n) = d\), then \(m = da\) and \(n = db\) where \(\gcd(a, b) = 1\). Substituting these into the divisibility condition, we get:
   \[
   k(db) + da \mid k(da) + db
   \]
   Simplifying, we have:
   \[
   d(kb + a) \mid d(ka + b)
   \]
   Since \(d\) is a common factor, we can divide both sides by \(d\):
   \[
   kb + a \mid ka + b
   \]

   Now, we need to show that \(kb + a \mid ka + b\). Consider the expression:
   \[
   ka + b = (kb + a) \cdot q + r
   \]
   where \(q\) is the quotient and \(r\) is the remainder. Since \(kb + a \mid ka + b\), the remainder \(r\) must be zero. Therefore:
   \[
   ka + b = (kb + a) \cdot q
   \]

   Rearranging, we get:
   \[
   ka + b - (kb + a) \cdot q = 0
   \]
   Simplifying further:
   \[
   ka + b - kbq - aq = 0
   \]
   \[
   ka + b = q(kb + a)
   \]

   Since \(\gcd(a, b) = 1\), the only way this can hold is if \(q = 1\) and \(kb + a = ka + b\). This implies:
   \[
   kb + a = ka + b
   \]
   \[
   (k-1)b = (k-1)a
   \]
   \[
   b = a
   \]

   Therefore, \(n = db\) and \(m = da\) imply \(n \mid m\).

   To find all nice numbers, we need to check the condition \(k > 2\). If \(k > 2\), then we choose \(a = 2k - 1\) and \(b = k - 2\):
   \[
   kb + a = k(k-2) + (2k-1) = k^2 - 2k + 2k - 1 = k^2 - 1
   \]
   \[
   k^2 - 1 \mid k^2 - 1
   \]
   \[
   k - 2 \mid 2k - 1
   \]
   \[
   k - 2 \mid 3
   \]
   \[
   k \in \{3, 5\}
   \]

   Therefore, the nice numbers are \(2, 3, 5\).

The final answer is \( \boxed{ 2, 3, 5 } \).