Prove That the Shortest Period of the Gregorian Calendar is 400 Years I ask for help.
I would like to resolve the following question:
Let $A$ be the set of natural numbers greater than 1582.
If, for every $Y\in A$ , the years $Y$ and $Y + n$ (of the actual calendar, the Gregorian) have the same calendar,
 then $n = 4k$, where $k$ is any positive integer.
Note:
Actually, what I want to prove is the following:
the shortest period of the Gregorian calendar is 400 years.
 A: Your claim is true, but a much stronger claim is true also.
The only $n$ such that any two years $n$ years apart have the same leap/nonleap status are those that are multiples of 400.
For an $n$ that is divisible by $4$ but not by $400$ you can find different years $n$ apart as one of the pairs $(1700,1700+n)$, $(1800, 1800+n)$, or $(1900,1900+n)$.
If $n$ is a multiple of $400$ it also happens to guaranteed that the two years start on the same day of the week.
However, if you take "same calendar" to also include the Easter falls on the same date in the two years (and revising the rules for Easter computations was a major part of the Gregorian reform), then $400$ will not do. If I understand the rules for the lunar computations correctly, the sequence of Gregorian Easter dates will only repeat exactly after $30\cdot\operatorname{gcm}(19,400,2500)=5{,}700{,}000$ years.
(How so? Brace yourself: Easter is the first Sunday after the first full moon after March 20, where "full moon" refers to an artificial cycle that tries to approximate the astronomical phenomena as they were known at Gregory XIII's time. The full-moon dates are derived from a number called the epact which is reckoned modulo 
$30$ and increases by $11$ each year, with occasional adjustments generated by from three different cycles. The "saltus lunae" adds $1$ to the epact every $19$ years; the "solar equation" subtracts $1$ from the epact three times in $400$ years; and the "lunar equation" adds $1$ eight times in $2500$ years. The combined sequence of adjustments to the epact will repeat after $\operatorname{gcm}(19,400,2500)=190{,}000$ years, but the total increase of the epact over this period happens to be coprime to $30$, so we need to multiply by $30$ before the epacts themselves repeat.)
A: I see different interpretations of your question:

There is $n$ such that $Y$ and $Y+n$ are the same for all $Y \ge 1583$. In this case, $n$ is a multiple of $4$.

Yes, $n=400$ is the smallest such $n$ and every such $n$ is a multiple of $400$.
On the other hand, $n=28$ works for all years between $1901$ and $2099$ because every fourth year after $1904$ is a leap year.

Given a year $Y$, if $Y$ and $Y+n$ are the same, then $n$ is a multiple of $4$.

No, $2007$ and $2018$ are the same and their difference is $11$.
The difference between years of the same type follows a complicated pattern of $5,6,11,28$, depending on the starting year.
A: Having the same calendar is synonymous with starting on the same day and in general each year starts a day later.  If there were no such thing as leap days,  $400 \equiv 1 \mod 7$, so a cycle of $400$ would be one day later.
If we had leap years every $4$ years then (as $4|400$) we'd have exactly $100$ more days and the calendar would be, $1 + 100 \equiv 3\mod 7$, $3$ laters.  But we don't have leap days every hundred years on years that are divisible by $100$.  As $100|400$ we'd have exactly $100-4$ leap days so the calender would be one day early.  ($1 + 96 \equiv 6 \mod 7$.)
But years divisible by $400$ do have a leap day and as $400|400$ we have exactly $97$ leap days.  So $400 + 97 \equiv 0 \mod 7$ and a cycle of $400$ years does return the calendar to start.
But for any number less than $400$ we can not guarantee an exact number of leap days.  For Example: the years from $1600$ to $1600+k$ will have one more leap year than the years fro $1700$ to $1700+ k$.  So no cycle less than $400$ is possible.
A: We will prove that there is no natural n < 400 such that for all Y, the Gregorian years Y and Y + n have identical calendars.
--- For the years 1600 and 1600 + n to have the same calendar, n must be multiple of 4.
--- If n is a multiple of 4, the years 1700 and 1700 + n, with n < 400, will only have the same calendar if n = 100 or n = 200.
--- If n = 100 or n = 200, the years 1600 and 1600 + n will have different calendars.
  So there is no n < 400, such that the years Y and Y + n have, for every Y, identical calendars.
A: For every 400 year period, there are four centurial years, three of which the leap year is ommitted, and on the fourth it is maintained. Hence, the minimal period of the Gregorian calendar must be greater than or equal to 400 years.
Now, the minimal period must have an integer value for the number of weeks in it.
We will now show that the minimal period is less than or equal to 400 years, upon which we may conclude the period is 400 years.
To this end, in any 400 year period, there are 303 weeks of 365 days and 97 weeks of 366 days. Hence, we solve the following congruence:
$$\left[(303)(365) + (97)(366)\right] \pmod7 \equiv 0 \pmod7.$$
Therefore, the minimal period of the Gregorian calendar is 400 years.
