jmc

Setting $m=n,$ we get $$f(nf(n))=nf(n).$$Thus, $nf(n)$ is a fixed point for all positive integers $n.$ (In other words, $x=nf(n)$ satisfies $f(x)=x.$)

Setting $m=1,$ we get $$f(f(n))=nf(1).$$If $n$ is a fixed point (which we know exists), then $n=nf(1),$ so $f(1)=1.$ Hence, $$f(f(n))=n$$for all positive integer $n.$ This equation tells us that the function $f$ is surjective.

Furthermore, if $f(a)=f(b),$ then $$f(f(a))=f(f(b)),$$so $a=b.$ Therefore, $f$ is injecitve, which means that $f$ is bijective.

Replacing $n$ with $f(n)$ in the given functional equation yields $$f(mf(f(n)))=f(n)f(m).$$Since $f(f(n))=n,$ $$f(mn)=f(n)f(m)(\ast )$$for all positive integers $m$ and $n.$

Taking $m=n=1$ in $(\ast ),$ we get $$f(1)=f(1{)}^2,$$so $f(1)=1.$

Recall that for a positive integer $n,$ $\tau (n)$ stands for the number of divisors of $n.$ Thus, given a positive integer $n,$ there are $\tau (n)$ ways to write it in the form $$n=ab,$$where $a$ and $b$ are positive integers. Then $$f(n)=f(ab)=f(a)f(b).$$Since$f$ is a bijection, each way of writing $n$ as the product of two positive integers gives us at least one way of writing $f(n)$ as the product of two positive integers, so $$\tau (f(n))\ge \tau (n).$$Replacing $n$ with $f(n),$ we get $$\tau (f(f(n))\ge \tau (f(n)).$$But $f(f(n))=n,$ so $$\tau (n)\ge \tau (f(n)).$$Therefore, $$\tau (f(n))=\tau (n)$$for all positive integers $n.$

If $n$ is a prime $p,$ then $$\tau (f(p))=\tau (p)=2.$$This means $f(p)$ is also prime. Hence, if $p$ is prime, then $f(p)$ is also prime.

Now, $$f(2007)=f(3^2\cdot 223)=f(3{)}^{2}f(223).$$We know that both $f(3)$ and $f(223)$ are prime.

If $f(3)=2,$ then $f(2)=3,$ so $f(223)\ge 5,$ and $$f(3{)}^{2}f(223)\ge 2^2\cdot 5=20.$$If $f(3)=3,$ then $$f(3{)}^{2}f(223)\ge 3^2\cdot 2=18.$$If $f(3)\ge 5,$ then $$f(3{)}^{2}f(223)\ge 5^2\cdot 2=50.$$So $f(2007)$ must be at least 18. To show that the 18 is the smallest possible value of $f(2007),$ we must construct a function where $f(2007)=18.$ Given a positive integer $n,$ take the prime factorization of $n$ and replace every instance of 2 with 223, and vice-versa (and all other prime factors are left alone). For example, $$f(2^7\cdot 3^4\cdot 223\cdot 11^5)=223^7\cdot 3^4\cdot 2\cdot 11^5.$$It can be shown that this function works. Thus, the smallest possible value of $f(2007)$ is $\boxed{18}.$