For example, $9 N_{0}$ 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 $f(1993)$.
(1994 Bulgarian Olympiad Problem)
First, we prove: If \( m, n \in \mathbb{N}_0 \) and \( m \neq n \), then \( f(m) \neq f(n) \), i.e., \( f \) is a one-to-one mapping.
By contradiction. If \( f(m) = f(n) \), then \( f[f(m)] = f[f(n)] \).
Thus, from the given condition, we have
\[
2m + 3 = f[f(m)] + f(m) = f[f(n)] + f(n) = 2n + 3,
\]
which implies \( m = n \). This is a contradiction.
Next, we find \( f(0) \). Let \( f(0) = x \) where \( x \geq 0 \).
In the given relation, let \( n = 0 \), then we get \( f(x) + x = 3 \), i.e., \( f(x) = 3 - x \).
Since \( f(x) \geq 0 \), it follows that \( x \leq 3 \).
From the given relation, let \( n = x \), then we get
\[
f[f(x)] + f(x) = 2x + 3.
\]
Substituting (1) into the above equation, we get \( f(3 - x) = 3x \).
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 equation, we get \( f(3x) = 9 - 5x \).
Since \( f(3x) \geq 0 \), it follows that \( 9 - 5x \geq 0 \), i.e., \( x \leq \frac{9}{5} \).
Thus, \( x = 0 \) or \( x = 1 \).
If \( x = 0 \), i.e., \( f(0) = 0 \). In the given relation, let \( n = 0 \), then we get \( f[f(0)] + f(0) = 3 \), which implies \( 0 = 3 \), a contradiction.
Therefore, \( x = 1 \), so \( f(0) = 1 \).
Using mathematical induction, we can prove that for any non-negative integer \( n \), \( f(n) = n + 1 \).
Indeed, when \( n = 0 \), by \( f(0) = 1 \), the conclusion holds.
Assume that when \( n = k \), the conclusion holds, i.e., \( f(k) = k + 1 \).
In the given relation, let \( n = k \), then we have \( f[f(k)] + f(k) = 2k + 3 \).
By the induction hypothesis, we get \( f(k + 1) + k + 1 = 2k + 3 \),
i.e., \( f(k + 1) = k + 2 \).
By the principle of mathematical induction, for any non-negative integer \( n \), \( f(n) = n + 1 \).
Thus, \( f(1993) = 1993 + 1 = 1994 \).