Freaky Function!

Note the function is N to N. Sentence 2 implies that $f(n-1) < f(n) < f(n+1)$ .

So: $0<1<2 \rightarrow f(0)=0$ and $f(1)=1$

Now, suppose you have two numbers that $f(k)=k$ and $f(k+1)=k+1$ Then:

$2k<2k+1<2k+2 \rightarrow 2f(k)<f(2k+1)<2f(k+1)$

$2k<f(2k+1)<2k+2 \rightarrow f(2k+1)=2k+1$ .

This means that $f(k)=k, f(k+1)=k+1 \rightarrow f(2k+1)=2k+1$ .

$f(1)=1, f(2)=2 \rightarrow f(3)=3, f(4)=4$

$f(2)=2 ,f(3)=3 \rightarrow f(5)=5, f(6)=6$

$f(3)=3, f(4)=4 \rightarrow f(7)=7, f(8)=8$ etc. Then $f(x)=x$ for all naturals.

By A.P., the answer will be $\frac{(1+20)20}{2}=210$ .