Suppose that $\{a_n\}$ is a sequence such that $a_{n+1}=(1+\frac{k}{n})a_{n}+1$ with $a_{1}=1$.Find all positive integers $k$ such that  any $a_n$ be integer.
1. We start with the recurrence relation given for the sequence $\{a_n\}$:
   \[
   a_{n+1} = \left(1 + \frac{k}{n}\right)a_n + 1
   \]
   with the initial condition $a_1 = 1$.

2. We need to find all positive integers $k$ such that $a_n$ is an integer for all $n$.

3. Let's first test the case $k = 1$:
   \[
   a_{n+1} = \left(1 + \frac{1}{n}\right)a_n + 1
   \]
   Starting with $a_1 = 1$:
   \[
   a_2 = \left(1 + \frac{1}{1}\right) \cdot 1 + 1 = 3
   \]
   \[
   a_3 = \left(1 + \frac{1}{2}\right) \cdot 3 + 1 = \frac{9}{2} + 1 = \frac{11}{2}
   \]
   Since $a_3$ is not an integer, $k = 1$ is not a solution.

4. Now, consider the case $k = 2$:
   \[
   a_{n+1} = \left(1 + \frac{2}{n}\right)a_n + 1
   \]
   Starting with $a_1 = 1$:
   \[
   a_2 = \left(1 + \frac{2}{1}\right) \cdot 1 + 1 = 4
   \]
   \[
   a_3 = \left(1 + \frac{2}{2}\right) \cdot 4 + 1 = 9
   \]
   \[
   a_4 = \left(1 + \frac{2}{3}\right) \cdot 9 + 1 = 15 + 1 = 16
   \]
   We observe that $a_n = n^2$ for $k = 2$. This can be verified by induction:
   - Base case: $a_1 = 1^2 = 1$
   - Inductive step: Assume $a_n = n^2$. Then,
     \[
     a_{n+1} = \left(1 + \frac{2}{n}\right)n^2 + 1 = n^2 + 2n + 1 = (n+1)^2
     \]
   Thus, $a_n = n^2$ for all $n$ when $k = 2$, and $a_n$ is always an integer.

5. For $k > 2$, we need to check if $a_{n+1}$ is always an integer. Using the given formula:
   \[
   a_{n+1} = \binom{n+k}{k} \frac{k}{k-1} - \frac{n+1}{k-1}
   \]
   We need to check if this expression is always an integer. For large $n$, the term $\frac{n+1}{k-1}$ may not be an integer, which implies that $a_{n+1}$ may not be an integer for some $n$.

6. Specifically, for $k > 2$, consider $n = (k-1)(k-1)! - 2$. In this case, $\frac{n+1}{k-1}$ is not an integer, which implies $a_{n+1}$ is not an integer.

Therefore, the only solution is $k = 2$.

The final answer is $\boxed{2}$.