Seven, let the sequence $\left\{a_{n}\right\}$ satisfy
$$
\begin{array}{l}
a_{1}=1, \\
a_{n+1}=\left(1+\frac{k}{n}\right) a_{n}+1(n=1,2, \cdots) .
\end{array}
$$

Find all positive integers $k$ such that every term in the sequence $\left\{a_{n}\right\}$ is an integer.
(Zhang Lei)
When $k=1$, $a_{2}=3, a_{3}=\frac{11}{2}$, which does not satisfy the condition.
When $k=2$, by the given condition we have
$$
\frac{a_{n+1}}{(n+1)(n+2)}=\frac{a_{n}}{n(n+1)}+\frac{1}{(n+1)(n+2)} \text {. }
$$

Thus, $\frac{a_{n}}{n(n+1)}=\frac{a_{1}}{1 \times 2}+\sum_{i=2}^{n} \frac{1}{i(i+1)}=1-\frac{1}{n+1}$.
Therefore, $a_{n}=n^{2}$ is a positive integer, satisfying the condition.
When $k \geqslant 3$, by the given condition we have
$$
\begin{array}{l}
\frac{a_{n+1}}{(n+1)(n+2) \cdots(n+k)} \\
=\frac{a_{n}}{n(n+1) \cdots(n+k-1)}+\frac{1}{(n+1)(n+2) \cdots(n+k)} .
\end{array}
$$

Thus, $\frac{a_{n}}{n(n+1) \cdots(n+k-1)}$
$$
\begin{array}{l}
=\frac{a_{1}}{1 \times 2 \times \cdots \times k}+\sum_{i=2}^{n} \frac{1}{i(i+1) \cdots(i+k-1)} \\
=\frac{1}{k-1}\left[\frac{1}{1 \times 2 \times \cdots \times(k-1)}-\frac{1}{(n+1)(n+2) \cdots(n+k-1)}\right] \\
\Rightarrow a_{n}=\frac{1}{k-1} \cdot \frac{n(n+1) \cdots(n+k-1)}{1 \times 2 \times \cdots \times(k-1)}-\frac{n}{k-1} \\
=\frac{(n+k-1)(n+k-2)}{(k-1)^{2}} \mathrm{C}_{n+k-3}^{k-2}-\frac{n}{k-1} .
\end{array}
$$

When $n+k-2=(k-1)^{2}, n=(k-1)^{2}-k+2$, by $\frac{(n+k-1)(n+k-2)}{(k-1)^{2}} \mathrm{C}_{n+k-3}^{k-2} \in \mathbf{N}, \frac{n}{k-1}$ $\notin \mathbf{N}$, we know
$$
a_{n}=\frac{(n+k-1)(n+k-2)}{(k-1)^{2}} \mathrm{C}_{n+k-3}^{k-2}-\frac{n}{k-1} \notin \mathbf{N} \text {. }
$$

In summary, the value of the positive integer $k$ that satisfies the condition is 2.