Let $f(n)$ be a function defined for all positive ...

We wish to evaluate

$$f(1)= \begin{cases} n-3,&\text{if }n\ge1000\text{ and }3\mid n,\\[1mm] f\bigl(f(n+5)\bigr),&\text{otherwise}, \end{cases}$$

at $n=1$. (Here “$3\mid n$” means “$n$ is divisible by 3”.)

Because $1<1000$, the “otherwise‐case” applies:

$$f(1)=f\bigl(f(1+5)\bigr)=f\bigl(f(6)\bigr).$$

In fact, if the input is less than 1000 or (even if $\ge1000$) is not divisible by 3, the definition tells us to “bump” the argument by 5 (inside two $f$’s). Thus, starting with $n=1$ we have

$$f(1)=f\bigl(f(6)\bigr),\quad f(6)=f\bigl(f(11)\bigr),\quad f(11)=f\bigl(f(16)\bigr),\quad\text{etc.}$$

The key observation is that if we repeatedly add 5 we eventually reach a number that is at least 1000 and divisible by 3 so that the first (non‐recursive) rule applies. In other words, if we define

$$N_k=1+5k,\qquad k=0,1,2,\dots,$$

we want the first $k$ for which

$$N_k\ge1000\quad\text{and}\quad3\mid N_k.$$

Finding the “Base‐case” Level

First, find the least $k$ such that

$$1+5k\ge1000\quad\Longrightarrow\quad k\ge \frac{999}{5}=199.8,$$

so $k\ge200$.

Next, we also require $1+5k$ to be divisible by 3. Notice that

$$1+5k\equiv1+2k\pmod3$$

(since $5\equiv2\pmod3$). We need

$$1+2k\equiv0\pmod3\quad\Longrightarrow\quad2k\equiv-1\equiv2\pmod3.$$

This is equivalent to

$$k\equiv1\pmod3.$$

The smallest $k\ge200$ that is congruent to 1 modulo 3 is $k=202$ (since $200\equiv2$ and $201\equiv0$ modulo 3).

Then

$$N_{202}=1+5\cdot202=1011.$$

Since $1011\ge1000$ and $1011$ is divisible by 3 (indeed, $1011=3\cdot337$), the definition tells us that

$$f(1011)=1011-3=1008.$$

Unwinding the Recursion

To “unwind” the nested calls we introduce the notation

$$g(k)=f(N_k)=f(1+5k).$$

Then the recursive definition tells us that for all $k$ for which $N_k$ is not “base‐case” we have

$$g(k)=f\bigl(f(1+5k+5)\bigr)=f\bigl(f(1+5(k+1))\bigr)=f\bigl(g(k+1)\bigr).$$

Thus, for $k$ below the level where the base case is used, the values “propagate” by the relation

$$g(k)=f\bigl(g(k+1)\bigr).$$

We already computed:

• $g(202)=f(1011)=1008$.

Now, let’s step back:

1. For $k=201$: $N_{201}=1+5\cdot201=1006$. Then

   $$g(201)=f(1006)=f\bigl(f(1006+5)\bigr)=f\bigl(f(1011)\bigr)=f(1008).$$

   But $1008\ge1000$ and $1008$ is divisible by 3, so

   $$f(1008)=1008-3=1005.$$

   Thus, $g(201)=1005$.

2. For $k=200$: $N_{200}=1+5\cdot200=1001$. Then

   $$g(200)=f(1001)=f\bigl(f(1001+5)\bigr)=f\bigl(f(1006)\bigr)=f(1005).$$

   And since $1005$ is divisible by 3,

   $$f(1005)=1005-3=1002.$$

   So, $g(200)=1002$.

3. For $k=199$: $N_{199}=1+5\cdot199=996$. Then

   $$g(199)=f(996)=f\bigl(f(996+5)\bigr)=f\bigl(f(1001)\bigr)=f(1002).$$

   Now, $1002\ge1000$ and is divisible by 3 so

   $$f(1002)=1002-3=999.$$

   Therefore, $g(199)=999$.

At this point a “two‐value cycle” appears. In fact, one may check that once the arguments drop below 1000 the recursion forces the following:

• $f(1002)=999$,
• $f(999)=1002$.

(One may verify, for example, that

$$f(999)=f\bigl(f(999+5)\bigr)=f\bigl(f(1004)\bigr),$$

and although $1004\ge1000$, it is not divisible by 3 so one must recurse further. A short computation shows that $f(1004)=1005$ and then $f(1005)=1002$. Hence, $f(999)=1002$.)

Thus, as we “back‐substitute” step by step from $k=202$ down to $k=0$ (where $g(0)=f(1)$) we see that the values alternate:

• $g(202)=1008$ (even “step”),
• $g(201)=1005$ (odd),
• $g(200)=1002$ (even),
• $g(199)=999$ (odd),
• $g(198)=1002$ (even),
• $g(197)=999$ (odd),
• $\dots$

Since the recursion “steps” down one unit of $k$ at a time and the alternation begins at $k=200$ (with $g(200)=1002$), we find that if we go back $200$ steps from $k=200$ to $k=0$, the parity of the number of steps (here, 200, which is even) will preserve the value. In other words, for $k$ with

$$200-k\text{ even}\quad\Longrightarrow\quad g(k)=1002,$$

and

$$200-k\text{ odd}\quad\Longrightarrow\quad g(k)=999.$$

Since $k=0$ gives $200-0=200$ even, we conclude that

$$g(0)=f(1)=1002.$$

Final Answer

Thus, the value of $f(1)$ is

$$\boxed{1002}.$$

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Summary of Key Steps

1. Recursive Definition: Notice that for inputs $n<1000$ or $n\ge1000$ not divisible by 3, the definition is $f(n)=f(f(n+5))$. Starting at $n=1$, repeated addition of 5 is required until a number $\ge1000$ and divisible by 3 is reached.

2. Finding the Base Case: By setting $1+5k\ge1000$ and $1+5k\equiv0\pmod3$, one finds that the smallest such $k$ is $202$, with $1+5\cdot202=1011$. Then $f(1011)=1011-3=1008$.

3. Backward Recursion and Alternation: Defining $g(k)=f(1+5k)$ gives the recurrence $g(k)=f(g(k+1))$. Unwinding the recursion from $k=202$ downwards shows that the values eventually alternate between $1002$ and $999$. In particular, $g(200)=1002$ and after an even number of backward steps we return to $1002$.

4. Conclusion: Since $g(0)=f(1)=1002$, the answer is $1002$.

The final answer is: $\boxed{1002}$.