Let $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}, a_{7}$ be non-negative numbers whose sum is 1. Let $M$ denote the maximum of the quantities $a_{1}+a_{2}+a_{3}, a_{2}+a_{3}+a_{4}, a_{3}+a_{4}+a_{5}$, $a_{4}+a_{5}+a_{6}, a_{5}+a_{6}+a_{7}$.

How small can $M$ be?
I. solution. The sum of the five trinomials obviously cannot be greater than $5 M$. After simplification, this sum can be written in the form

$$
S=3\left(a_{1}+a_{2}+\ldots+a_{7}\right)-a_{1}-\left(a_{1}+a_{2}\right)-\left(a_{6}+a_{7}\right)-a_{7}
$$

Since our numbers are non-negative, none of the subtrahends can be greater than $M$, so $S \geqq 3-4 M$.

From the condition $5 M \geqq S \geqq 3-4 M$, we get $M \geqq \frac{1}{3}$. For equality, $a_{1}=a_{1}+a_{2}=a_{6}+a_{7}=a_{7}=\frac{1}{3}$, which means $a_{1}=a_{7}=\frac{1}{3}, a_{2}=a_{6}=0$ is necessary. If we choose the values of the remaining three numbers such that $a_{3}$ and $a_{5}$ are 0, and $a_{4}$ is also $\frac{1}{3}$, then the sum of these seven numbers is 1, and the value of each of the five trinomials, and thus the value of $M$, is $\frac{1}{3}$.

Therefore, the smallest value of $M$ is $\frac{1}{3}$.

II. solution. None of the five trinomials can be greater than $M$, so by adding the first, fourth, and fifth of them, we cannot get a value greater than $3 M$. After simplification, we get

$$
\left(a_{1}+a_{2}+\ldots+a_{7}\right)+\left(a_{5}+a_{6}\right) \leqq 3 M
$$

from which, using the fact that the sum of the seven numbers is 1, and that $a_{5}+a_{6} \geqq 0$,

$$
1 \leqq 3 M, \quad \text { i.e., } \quad \frac{1}{3} \leqq M
$$

follows. For equality, now $a_{5}=a_{6}=0$ is necessary, from which, proceeding further, we get the seven values obtained in the first solution.

János Csirik (Szeged, Ságvári E. Gymnasium, 1st year)

Note. The following generalization of the problem also comes from János Csirik. If the sum of $n$ non-negative numbers is 1, the number of addends in the sums of adjacent elements is $k(0<k \leqq n)$, and we are looking for the minimum of the maximum $M$ of the sums $a_{1}+a_{2}+\ldots+a_{k}, a_{2}+\ldots+a_{k+1}, \ldots$, $a_{n-k+1}+\ldots+a_{n}$, then $M=\frac{1}{\left\lceil\frac{n}{k}\right\rceil}$, where $\lceil x\rceil$ is the "upper integer part" of $x$, i.e., the smallest integer not less than $x$. (In the problem, $n=7, k=3,\left\lceil\frac{n}{k}\right\rceil=3$.) If $m=\left\lceil\frac{n}{k}\right\rceil$, then $M=\frac{1}{m}$, for example, if $a_{1}=a_{k+1}=\ldots=a_{(m-1) k+1}=\frac{1}{m}$, and the other $a_{i}$ are 0.