At the "Marslakók," a year lasts 687 days. Since the Marslakó memory, months have been 26 to 29 days long. The Renewers of All propose to switch to months of 27 and 31 days. With the support of the Nyírbálók, they manage to get this accepted, hoping that this will reduce the number of months (and thus wages). Is the idea of the Nyírbálók feasible?
A 687-day "Mars-year" consists of $x$ 26-day months and $y$ 29-day months, while the Nyírbálók propose to change this to $p$ 27-day months and $q$ 31-day months, then with suitable positive integers $x, y, p, q$

$$
26 x+29 y=687 \quad \text { and }(1) 27 p+31 q=687, \quad \text { furthermore }(2)
$$

According to the Nyírbálók's intention, $p+q<x+y$.

From (1), $26 x+26 y<687<29 x+29 y$, hence

$$
23<\frac{687}{29}<x+y<\frac{687}{26}<27
$$

The possible values of $x+y$ are therefore 24, 25, and 26.

Similarly, from (2), we get $27 p+27 q<687<31 p+31 q$, that is,

$$
22<\frac{687}{31}<p+q<\frac{687}{27}<26
$$

Thus, the possible values of $p+q$ are 23, 24, and 25.

The grand reform, therefore, seems feasible, but let's consider that $x, y, p$, and $q$ are all integers.

Adding $(x+y)$ to both sides of (1), and $(p+q)$ to both sides of (2),

$$
27 x+30 y=687+x+y, (3) 28 p+32 q=687+p+q \cdot (4)
$$

The left side of (3) is divisible by 3, and the left side of (4) is divisible by 4. Since 687 is divisible by 3 and leaves a remainder of 3 when divided by 4, $x+y$ must be divisible by 3, and $p+q$ must leave a remainder of 1 when divided by 4.

Among the possible values, this is only true for $x+y=24$ and $p+q=25$, despite the shorter months.

The Nyírbálók's attempt, therefore, cannot succeed; the number of months does not decrease but increases after the reform.

Note. For completeness, it can be verified that with the choice of $x+y=24$ and $p+q=25$, there indeed exists a division of the 687-day year into months of the given lengths. (In the solution, we only used the necessary conditions for such divisions.)

If $x+y=24$, then $3 y=687-26(x+y)=63$, so $y=21$ and $x=3$, and if $p+q=25$, then $4 q=687-27(p+q)=12$, that is, $q=3$ and $p=22$.