17. Let $N=\{0,1,2, \cdots\}$, given a $k \in \mathbf{N}$. Try to find all functions $f: \mathbf{N} \rightarrow \mathbf{N}$, such that for every $n \in \mathbf{N}$, $f(n)<f(n+1)$, and $f(f(n))=n+2k$.
17. First prove the following lemma.

Lemma If the function $f: \mathbf{N} \rightarrow \mathbf{N}$ satisfies $f(n)<f(n+1)$ for every $n \in \mathbf{N}$, then for any $m, n \in \mathbf{N}, m \geqslant n$, it must be true that $f(m)-f(n) \geqslant m-n$, and $f(m) \geqslant m$.
Proof of the lemma: Since $f(n)<f(n+1)$, and $f(n), f(n+1)$ are non-negative integers, we have
$$
\begin{array}{l}
f(n+1)-f(n) \geqslant 1, \\
f(n+2)-f(n+1) \geqslant 1, \\
\cdots \cdots \cdot \\
f(m)-f(m-1) \geqslant 1 .
\end{array}
$$

Adding the above $m-n$ inequalities, we get
$$
f(m)-f(n) \geqslant m-n .
$$

Taking $n=0$ in the above equation, we get
$$
f(m) \geqslant m+f(0) \geqslant m .
$$

Now, we prove the original problem.
From $f(f(n))=n+2k$ and the lemma, we have
$$
n+2k-f(n)=f(f(n))-f(n) \geqslant f(n)-n,
$$

which means
$$
f(n) \leqslant n+k \quad (n \in \mathbf{N}) \text{. }
$$

In equation (1), replace $n$ with $f(n)$ to get
$$
n+2k=f(f(n)) \leqslant f(n)+k,
$$

which means
$$
f(n) \geqslant n+k \quad (n \in \mathbf{N}) .
$$

From equations (1) and (2), we get $f(n)=n+k \quad (n \in \mathbf{N})$.
Clearly, this function satisfies the problem's requirements.
In conclusion, the only function that meets the criteria is $f(n)=n+k \quad (n \in \mathbf{N})$.