5. For a set $M$ containing 4 elements, we say it is "evenly connected" if for every element $x$ in $M$, at least one of the numbers $x-2$ or $x+2$ also belongs to $M$. Let $S_{n}$ be the number of "evenly connected" subsets of $\{1,2, \ldots, n\}$. Determine the smallest number $n$ such that $S_{n} \geq 2011$.
Solution. Let $\mathbf{M}=\{a, b, c, d\}$ be a "pairwise connected" set. Without loss of generality, we can assume that $a < b < c < d$. This form will be called form 1.

iii) $b=a+2 ; a=d-2$. This case is not possible because it results in $b=d$.

iv) $c=a+2 ; b=d-2$. It follows that $b=a+1$ and $d=a+3$, and the set $\mathbf{M}$ is of the form $\{a, a+1, a+2, a+3\}$. This form will be called form 2.

v) $c=a+2 ; c=d-2$. It follows that $b=a+1, d=a+4$ and the set $\mathbf{M}$ is of the form $\{a, a+1, a+2, a+4\}$, but $a-1, a+3 \notin \mathbf{M}$. Thus, this case is not possible.

vi) $c=a+2 ; a=d-2$. This case is not possible because it results in $c=d$.

vii) The case $d=a+2$ is not possible. With this, all possibilities for the form of $\mathbf{M}$ are exhausted.

Let $n$ be sufficiently large. For $\{1,2, \ldots, n\}$, "pairwise connected" subsets of form 2 are a total of $n-3$. It remains to calculate the number of "pairwise connected" subsets of form 1.

If $a=1$ there are a total of $n-5$ different subsets,

If $a=2$ there are a total of $n-6$ different subsets,

If $a=3$ there are a total of $n-7$ different subsets,

If $a=n-5$ there is only one such subset.

Thus, the total number of "pairwise connected" subsets of form 1 is

$$
1+2+\ldots+(n-5)=\frac{(n-5)(n-4)}{2}=\frac{n^{2}-9 n+20}{2}
$$

Thus, the total number of "pairwise connected" subsets of $\{1,2, \ldots, n\}$ is:

$$
S_{n}=n-3+\frac{n^{2}-9 n+20}{2}=\frac{n^{2}-7 n+14}{2}=\frac{(n-3)(n-4)}{2}+1
$$

We need to determine the smallest $n$ for which $\frac{(n-3)(n-4)}{2}+1 \geq 2011$, or $(n-3)(n-4) \geq 4020$. By checking, it is confirmed that $n \geq 67$. Thus, the smallest number is 67.