6.26 $N_{0}$ is the set of all non-negative integers, $f(n)$ is a function such that $f: N_{0}$
$\rightarrow N_{0}$, and for each $n \in N_{0}, f(f(n))+f(n)=2 n+3$. Find: $f(1993)$.
[ Sol] First, we prove that if $m, n \in N_{0}, m \neq n$, then $f(m) \neq f(n)$, i.e., $f$ is $1-1$.

By contradiction.
If $f(m)=f(n)$, then $f(f(m))=f(f(n))$. Thus, by the given condition, we have
$$2 m+3=f(f(m))+f(m)=f(f(n))+f(n)=2 n+3$$

Hence, $m=n$, a contradiction.
Next, find $f(0)$.
Let $f(0)=x, x \geqslant 0$.
In the given relation, let $n=0$, then we get
$$\begin{array}{l}
f(x)+x=3 \\
f(x)=3-x
\end{array}$$

Since $f(x) \geqslant 0$, it follows that $x \leqslant 3$.
In the given relation, let $n=x$, then we get
$$f(f(x))+f(x)=2 x+3$$

Substituting (1) into the above, we get
$$f(3-x)=3 x$$

In the given relation, let $n=3-x$, then we get
$$f(f(3-x))+f(3-x)=2(3-x)+3$$

Substituting (2) into the above, we get
$$f(3 x)=9-5 x$$

Since $f(3 x) \geqslant 0$, it follows that
$$\begin{array}{l}
9-5 x \geqslant 0 \\
x \leqslant \frac{9}{5}
\end{array}$$

Thus, $x=0$ or $x=1$.
If $x=0$, i.e., $f(0)=0$. In this case, in the given relation, let $n=0$, then we get
$$f(f(0))+f(0)=3$$

Thus, $0=3$, a contradiction.
Therefore, it must be that $x=1$, so $f(0)=1$.
Next, we use mathematical induction to prove that for any non-negative integer $n, f(n)=n+1$.
In fact, when $n=0$, by $f(0)=1$, the conclusion holds.
Assume that when $n=k$, the conclusion holds, i.e., $f(k)=k+1$.
Then, in the given relation, let $n=k$, we have
$$f(f(k))+f(k)=2 k+3$$

By the induction hypothesis, we get
$$\begin{array}{l}
f(k+1)+k+1=2 k+3 \\
f(k+1)=k+2
\end{array}$$

According to the principle of mathematical induction, for any non-negative integer $n$, we have
$$f(n)=n+1$$

Therefore, the required $f(1993)=1993+1=1994$.