Consider a sequence of real numbers of finite length. Consecutive four term averages of this sequence are strictly increasing, but consecutive seven term averages are strictly decreasing. What is the maximum possible length of such a sequence?

Let the finite sequence be $$a_1,a_2,\dots ,a_n$$.

Denote the sliding four-term averages by $$S_k=\dfrac{a_k+a_{k+1}+a_{k+2}+a_{k+3}}{4},\;1\le k\le n-3$$.
Because these averages are strictly increasing,

$$S_{k+1}\gt S_k \;\Longrightarrow\; a_{k+1}+a_{k+2}+a_{k+3}+a_{k+4}\gt a_k+a_{k+1}+a_{k+2}+a_{k+3} \;\Longrightarrow\; a_{k+4}\gt a_k \quad -(1)$$

Similarly, let the seven-term averages be $$T_k=\dfrac{a_k+a_{k+1}+\dots +a_{k+6}}{7},\;1\le k\le n-6.$$ Because these are strictly decreasing,

$$T_{k+1}\lt T_k \;\Longrightarrow\; a_{k+1}+a_{k+2}+\dots +a_{k+7}\lt a_k+a_{k+1}+\dots +a_{k+6} \;\Longrightarrow\; a_{k+7}\lt a_k \quad -(2)$$

Thus, for every admissible index $$k$$ we have the pair of inequalities

$$a_{k+4}\gt a_k,\qquad a_{k+7}\lt a_k.$$

Case 1: $$n\ge 11$$ (we shall derive a contradiction).

Consider the eleven indices in the order
$$1,\;8,\;4,\;11,\;7,\;3,\;10,\;6,\;2,\;9,\;5.$$

Using (1) and (2) successively we obtain the strict chain

$$\begin{aligned} a_1 &\gt a_8 &&\bigl(\text{from }(2)\text{ with }k=1\bigr)\\ a_8 &\gt a_4 &&\bigl((1)\text{ with }k=4\bigr)\\ a_4 &\gt a_{11} &&\bigl((2)\text{ with }k=4\bigr)\\ a_{11}&\gt a_7 &&\bigl((1)\text{ with }k=7\bigr)\\ a_7 &\gt a_3 &&\bigl((1)\text{ with }k=3\bigr)\\ a_3 &\gt a_{10} &&\bigl((2)\text{ with }k=3\bigr)\\ a_{10}&\gt a_6 &&\bigl((1)\text{ with }k=6\bigr)\\ a_6 &\gt a_2 &&\bigl((1)\text{ with }k=2\bigr)\\ a_2 &\gt a_9 &&\bigl((2)\text{ with }k=2\bigr)\\ a_9 &\gt a_5 &&\bigl((1)\text{ with }k=5\bigr) \end{aligned}$$

Finally, applying (1) with $$k=1$$ gives $$a_5\gt a_1$$, so altogether we have

$$a_1\gt a_8\gt a_4\gt a_{11}\gt a_7\gt a_3\gt a_{10}\gt a_6\gt a_2\gt a_9\gt a_5\gt a_1,$$

a strict cycle of inequalities that is impossible for real numbers. Hence no sequence can satisfy the given conditions when $$n\ge 11$$, and therefore

$$n\le 10.$$

Case 2: Construction of a sequence of length $$10$$ that fulfils the requirements.

Take

$$a_1=10,\; a_2=20,\; a_3=30,\; a_4=0,\; a_5=12,\; a_6=22,\; a_7=35,\; a_8=8,\; a_9=15,\; a_{10}=25.$$

Check the nine inequalities that must hold:

Four-term type (1):
$$a_5\gt a_1\;(12\gt10),\; a_6\gt a_2\;(22\gt20),\; a_7\gt a_3\;(35\gt30),\; a_8\gt a_4\;(8\gt0),$$ $$a_9\gt a_5\;(15\gt12),\; a_{10}\gt a_6\;(25\gt22).$$

Seven-term type (2):
$$a_8\lt a_1\;(8\lt10),\; a_9\lt a_2\;(15\lt20),\; a_{10}\lt a_3\;(25\lt30).$$

All conditions are satisfied, so a length-10 sequence exists.

Combining Case 1 and Case 2, the maximal possible length of the sequence is

10.