This is in my Top 10 of all IMO problems. Not only because I was a competitor then, but also because I was present to the "problem solving session" organized by the hosts and I wached the Czech student Martin Chadek presenting his solution, which made the jury K.O. Here it is: Let $x_1,x_2,\ldots$ be the given sequence and let $s_n=x_1+x_2+\ldots+x_n$. The conditions from the hypothesis can be now written as $s_{n+7}<s_n$ and $s_{n+11}>s_n$ for all $n\ge 1$. We then have: $0<s_{11}<s_4<s_{15}<s_8<s_1<s_{12}<s_5<s_{16}<s_9<s_2<s_{13}<s_6<s_{17}<s_{10}<s_3<s_{14}<s_7<0,$ a contradiction. Therefore, the sequence cannot have $17$ terms. In order to show that $16$ is the answer, just take 16 real numbers satisfying $s_{10}<s_3<s_{14}<s_7<0<s_{11}<s_4<s_{15}<s_8<s_1<s_{12}<s_5<s_{16}<s_9<s_2<s_{13}<s_6$. We have $x_1=s_1$ and $x_n=s_n-s_{n-1}$ for $n\ge 2$. Thus we found all sequences with the given properties. As far as I know, Martin only solved this problem and got a special prize for it Other solutions are possible, including one using linear algebra. I'll post them if anyone is interested.

Suppose a sequence of $17$ terms exists. Consider the matrix $$\begin{bmatrix} x_1 & x_2 & x_3 & x_4 & x_5 & x_6 & x_7 & x_8 & x_9 & x_{10} \ \ \ x_{11} \\ x_2 & x_3 & x_4 & x_5 & x_6 & x_7 & x_8 & x_9 & x_{10} & x_{11} \ \ \ x_{12} \\ x_3 & x_4 & x_5 & x_6 & x_7 & x_8 & x_9 & x_{10} & x_{11} & x_{12} \ \ \ x_{13} \\ x_4 & x_5 & x_6 & x_7 & x_8 & x_9 & x_{10} & x_{11} & x_{12} & x_{13} \ \ \ x_{14} \\ x_5 & x_6 & x_7 & x_8 & x_9 & x_{10} & x_{11} & x_{12} & x_{13} & x_{14} \ \ \ x_{15} \\ x_6 & x_7 & x_8 & x_9 & x_{10} & x_{11} & x_{12} & x_{13} & x_{14} & x_{15} \ \ \ x_{16} \\ x_7 & x_8 & x_9 & x_{10} & x_{11} & x_{12} & x_{13} & x_{14} & x_{15} & x_{16} \ \ \ x_{17} \end{bmatrix}.$$ The sum of the numbers in each of the rows is positive. The sum of the numbers in each of the columns is negative. That is a contradiction. It is easy to show that $16$ works

Arne wrote: It is easy to show that 16 works This is the jury solution. They provided the example $$5,5,-13,5,5,5,-13,5,5,-13,5,5,5,-13,5,5.$$ But how do you find such an example? That's why I loved Martin's solution

More generally, it can be shown that : If $p,q$ are any two distinct positive integers such that none of them divides the other, and if a sequence has the property that the sum of any $p$ consecutive terms is always positive and the sum of any $q$ consecutive terms is negative then the maximal length of this sequence is $p+q-d-1$ where $d = \gcd (p,q).$ The proof I know is quite long, so that I will not write it here. It can be found in : Quadrature, n°54 (Oct. Dec. 2004), ex. E-218, p.34-36. Pierre.

Well, this was proved during the competition by GB7, John Rickard, a gold medalist in 1977!

enescu wrote: As far as I know, Martin only solved this problem and got a special prize for it No, he also fully solved Problem 4.

Does this work or have I deluded myself? Not fully detailed. We claim that the answer is $16$. One can find a construction without too much difficulty. Call an $n$-sum the sum of $n$ consecutive terms. If we suppose, for contradiction, that the sequence has $17$ terms, then every $4$-sum can be written as an $11$-sum with a $7$-sum subtracted, and must hence be positive. Similarly, every $3$-sum can be written as a $7$-sum with a $4$-sum subtracted, and must therefore be negative. And, finally, every term can be written as a $4$-sum with a $3$-sum subtracted, and must therefore be positive again. However, if every term is positive, then it's impossible for every $7$-sum to be negative, and we've reached a contradiction.

enescu wrote: Arne wrote: It is easy to show that 16 works This is the jury solution. They provided the example $$5,5,-13,5,5,5,-13,5,5,-13,5,5,5,-13,5,5.$$ But how do you find such an example? That's why I loved Martin's solution >16 years late, but I think you can motivate this solution by noticing that $\frac 27>\frac 3{11}$, meaning that if we can make it so that we get exactly two negative numbers of pretty large magnitude in each block of length $7$ and $3$ in each block of length $11$ we would be done by making the other numbers all equal to some positive number. It's not so obvious to pick the numbers $5$ and $-13$, but maybe this is easier? $$20, 20, -51, 20, 20, 20, -51, 20, 20, -51, 20, 20, 20, -51, 20, 20.$$ Note that the block of $11$ reaches from the first $-51$ to right before the last $20$.

fairly popular technique in competitive programming - prefix sums! The answer is $16$. For $1 \leq |S| \leq 16$, we can consider any subset of $\{5,5,-13,5,5,5,-13,5,5,-13,5,5,5,-13,5,5\}$. For $0 \leq k \leq 17$, then suppose $P_k = S_1 + S_2 + \cdots + S_k$. Then, $P_{k} < P_{k-7}$ and $P_{k} < P_{k + 11}$. Therefore, $$\begin{align*} P_{0} < P_{11} < P_4 < P_{15} < P_1 < P_{12} < P_{5} < P_{16} < P_2 < P_{13} < P_6 < P_{17} < P_3 < P_{14} < P_0, \end{align*}$$ which is clearly not possible. So, any sequence of $17$ real numbers has seven consecutive terms which sum to a negative value and has eleven consecutive terms which sum to a positive value. Thus, the maximum number of terms in the sequence is $\boxed{16}$.

If $n \geq 17$ then take the first 17 numbers $a_1, \ldots, a_{17}$ and write them as follows. $$\begin{bmatrix} a_1 & a_2 & a_3 & a_4 & a_5 & a_6 & a_7\\ a_2 & \ddots & & & & & a_8 \\ \vdots & &\ddots & & & & \vdots\\ \vdots & & & \ddots & &\\ a_{11} & & & & & &a_{17} \end{bmatrix}$$ Summing across the rows we note that the total sum is negative, while summing across the columns we get that the total sum is positive. That is a contradiction. Thus $n \leq 16$, and $n=16$ can be constructed (posted in other posts above).