Let's apply your strategy of using the Inclusion-Exclusion Principle.
There are $\binom{10}{4}$ four-element subsets of a ten-element set. From these, we must exclude those subsets containing at least two consecutive numbers.
A string of at least two consecutive numbers in the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ can begin with any of the nine numbers less than $10$. Once we select a pair of consecutive numbers beginning with one of these nine numbers, we must select two additional elements of the ten-element set from the eight elements that are not in the pair of consecutive numbers, which we can do in $\binom{8}{2}$ ways. Thus, it appears we have $9 \cdot \binom{8}{2}$ sets containing at least two consecutive numbers. However, we have overcounted since we have counted sets containing two disjoint pairs twice. There are $\binom{8}{2}$ such sets since we must exclude from the $\binom{9}{2}$ ways of selecting two of the nine numbers less than $10$ as starting points for the two pairs of consecutive integers the $\binom{8}{1}$ ways of selecting consecutive starting points that would prevent them from being disjoint. Note that it is a consequence of Pascal's Identity that $\binom{9}{2} - \binom{8}{1} = \binom{8}{2}$. Hence, there are
$$9 \cdot \binom{8}{2} - \left[\binom{9}{2} - \binom{8}{1}\right] = 9 \cdot \binom{8}{2} - \binom{8}{2} = 8 \cdot \binom{8}{2}$$
four-element subsets containing at least two consecutive numbers.
A string of at least three consecutive numbers in the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ can begin with any of the eight numbers less than $9$. Once we select a triple of consecutive numbers, we can select the remaining element in the subset from one of the seven elements in the set that are not in the triple, giving
$$8 \cdot \binom{7}{1}$$
four-element subsets that contain at least three consecutive numbers.
A string of four consecutive numbers in the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ can begin with any of the seven numbers less than $8$.
Thus, by the Inclusion-Exclusion Principle, the number of four-element subsets of the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$ that do not contain consecutive numbers is
$$\binom{10}{4} - 8 \cdot \binom{8}{2} + 8 \cdot \binom{7}{1} - 7 = 210 - 224 + 56 - 7 = 35$$
Addendum: A more efficient approach is to arrange six blue and four green balls in a row so that no two of the green balls are consecutive, then number the balls from left to right. The numbers on the green balls are the desired subset of four numbers, no two of which are consecutive, selected from the set $[10] = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}$.
Line up six blue balls in a row. This creates seven spaces, five between successive blue balls and two at the ends of the row.
$$\square b \square b \square b \square b \square b \square b \square$$
To ensure that no two of the green balls are consecutive, we select four of these seven spaces in which to place a green ball, which can be done in
$$\binom{7}{4} = 35$$
ways. We now number the balls from left to right. The numbers on the green balls are the desired subset of four numbers selected from the set $[10]$, no two of which are consecutive.
Let $S = \{1, 2, 3, \ldots, 3n - 2, 3n - 1, 3n\}$. Let
\begin{align*}
A & = \{k \in S \mid k \equiv 0 \pmod{3}\}\\
B & = \{k \in S \mid k \equiv 1 \pmod{3}\}\\
C & = \{k \in S \mid k \equiv 2 \pmod{3}\}
\end{align*}
Observe that $|A| = |B| = |C| = n$. We can choose three numbers from $S$ in the following ways:
- Choose $3$ elements of $A$, which can be done in $\binom{n}{3}$ ways.
- Choose $3$ elements of $B$, which can be done in $\binom{n}{3}$ ways.
- Choose $3$ elements of $C$, which can be done in $\binom{n}{3}$ ways.
- Choose $1$ element from each subset, which can be done in $\binom{n}{1}^3$ ways.
Hence, the number of ways of selecting a subset of three numbers in $S$ whose sum is divisible by $3$ is
\begin{align*}
3\binom{n}{3} + \binom{n}{1}^3 & = 3 \frac{n(n - 1)(n - 2)}{3!} + n^3\\
& = \frac{n^3 - 3n^2 + 2n}{2} + n^3\\
& = \frac{3n^3 - 3n^2 + 2n}{2}
\end{align*}
Best Answer
As Nate says in the comments: the sets where both $a$ and $b$ are divisible by 5 are counted twice as both numbers can play the role of $a$.
In order to correct for this we simply have to subtract off the number of these double counts which is the number of pairs $\{a,b\}$ with $a$ and $b$ divisible by 5 which is $$ \binom{200}{2} = 19900 $$ as you have noticed that there are 200 elements divisible by 5. When we have done this the answers then do agree and so both approaches are perfectly reasonable.