10. (1990 Japan Olympiad Selection Test) The function $f(n)$ with domain of positive integers satisfies: $f(n)=\left\{\begin{array}{l}n-3 \quad(n \geqslant 1000) \\ f[f(n+7)](n<1000)\end{array}\right.$, then $f(90)=$ $\qquad$ .
10. At first glance, the condition and the value to be found seem unrelated, making it difficult to start. By considering the relationship between 90 and 1000, we need to set up an auxiliary function to explore the pattern and use the periodicity of the function to solve the problem.
Let \( f_{m}(n) = \underbrace{f(f \cdots f}_{m \text{ times}}(n)) \).
Given that \( 90 + 7 \times 130 = 1000 \).
Thus, \( f(90) = f_{2}(97) = \cdots = f_{131}(1000) \).
Also, since \( f(1000) = 997 \).
Therefore, \( f(997) = f_{2}(1004) = f(1001) = 998 \),
\( f(998) = f_{2}(1005) = f(1002) = 999 \).
\( f(999) = f_{:}(1006) = f(1003) = 1000 \).
Thus, \( f_{m}(1000) \) is a periodic function with a period of 4.
Given that \( 131 = 4 \times 32 + 3 \).
Therefore, \( f(90) = f_{131}(1000) = f_{3}(1000) = 999 \).