If an infinite number of such $K$ do not exist, there must exist some $K_o$, such that for all $K > K_o$ there exists an $n$ with $n \le K < f(n)$. Since $|f(x) - x| \le 3$ must always hold, such an $n$ can only be $K$, $K-1$ or $K-2$.

The same must hold for $K = f(n)$, and so on. This means that, from some $K_o$ on, the function must map “up” into each interval $[n, f(n)]$, and also “up” out of each such interval. In order for this to be possible, we must have $f(n) - n \ge 3$, and it therefore follows that $|f(n) - n| = 3$ must hold for all $n > K > K_o$. If this is the case, we have
$$M(n) = \frac{1}{n+1} \sum_{j=0}^{n} |f(j) - j| = \frac{1}{n+1} \left( \sum_{j=0}^{K} |f(j) - j| + 3(n-k) \right) = 3 - \frac{C}{n+1}$$
with
$$C = 3K + 3 - \sum_{j=0}^{K} |f(j) - j| = \sum_{j=0}^{K} (3 - |f(j) - j|) \ge 0.$$
It therefore follows that there exists a $K_1$ such that $M(n) = 3 - \frac{C}{n+1} > 2.011$ for $n > K_1$, which contradicts the assumption $M(n) < 2.011$. This is not possible, and we see that an infinite number of $K$ with the required properties exist, as claimed.