We can see, for example, that the years 2009 and 2015 have identical calendars. Similarly, 2000 and 2028.
I read once that given any year X at most 28 years later there will be another year Y, with the calendar identical to that of X.
Here I am referring to our usual calendar, the Gregorian.
I have not yet been able to prove such an assertion. I ask for help.
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As José points out, if there were no irregularities in leap years, the calendar would repeat every 28 years, so there certainly will be another year with the same calendar within the next 28 (the 28th, if not before). In fact, it is only the leap years which take that long: if you started from a non-leap year, assuming no irregularities in leap years, you would get another year with the same calendar within $11$ years (it is $6$ or $11$ years later, depending on the year).
So the only years we need to worry about are when there is a skipped leap year coming up. If the year we start from, $X$, is not a multiple of $4$, then we are ok unless the skipped leap year occurs within the next $11$ years. But if it does, the $12$ years starting from $X$ contain $10$ years of $365$ days and $2$ of $366$. Since $10\times 365+2\times 366$ is a multiple of $7$, $X+12$ will have the same calendar as $X$.
If $X$ is a skipped leap year, then $X+6$ will have the same calendar.
If $X$ is a leap year, and one of the next two leap years is skipped, then again $X+12$ will start $10\times 365+2\times 366$ days later, and it will also be a leap year, so will have the same calendar as $X$. However, you can check that if a leap year in $X+12,...,X+28$ is skipped, the next year with the same calendar as $X$ will be more than $28$ years later.
So the only years which don't work are years ending 72, 76, 80, 84 or 88, where the next century year is not a leap year. This is $15$ failures in every $400$-year period.
That's because: