Combined 100 year birthday I was recently adding up the ages of myself and my wife and kids and realised that at some point in the not-too-distant future our ages will all add up to exactly 100 years.
Challenge
For an input consisting of a combined (total) age and a list of birth dates, output the date on which the combined ages all add up to the given total.

Input combined age (in years) will be a positive integer
Input list of birth dates will be a list (whatever format is convenient for your language) of dates that must contain numerical representation of Day, Month and Year.  Number of days since a specific epoch is not acceptable.  You may assume the list of input dates is sorted in chronological order.
Output will be a single date in the same format as input dates
For the purposes of the combined age, 1 year is considered to be exactly 365.25 days
For some inputs, it will be impossible to find a date after all the birth dates when they all add up to the combined age.  For instance consider two birth dates that are 20 years apart, but we want a combined of 10 years.  In that case, the output would be the 10th birthday of the older birth date.  In other words, individual ages are considered to be 0 for all dates before that individual's birth date
The output will be the first date on which the combined ages add up to at least the input age
You may use built-in date-time functions
You must support dates going back as far as 1970/01/01.

Examples
Here I'm giving all dates in YYYY/MM/DD format, but you may whatever format you like.
Input                                               Output

 10  2000/01/01                                     2010/01/01
100  1975/03/05,1978/07/23,2008/11/12,2012/12/20    2018/11/22
100  1975/03/06,1978/07/23,2008/11/12,2012/12/20    2018/11/23
100  1975/03/09,1978/07/23,2008/11/12,2012/12/20    2018/11/23
100  1975/03/10,1978/07/23,2008/11/12,2012/12/20    2018/11/24
  3  1990/08/01,1995/07/02                          1993/08/01
 50  1975/03/05,1978/07/23,2008/11/12,2012/12/20    2001/11/13

Note that in the last two examples, the output date lies within the range of input dates, as per rule 5.
Mathematica 138 107 237 bytes
My first attempts were sloppy and unworthy. This should work generally, provided that the birth dates are given in order of earliest to latest, as in the test cases.
The test cases were well-chosen and difficult to implement correctly.
The code turned out way longer than I had hoped.
The general logic is to

Solve for x (in days) such that the distance from x to each date of birth sums to the target passage of time (e.g 100 years converted to days).
If the earliest date + x days &gt; latest date of birth, remove the latest date of birth and repeat step (1), Else move to (3).
Return (earliest date + x days)


t_~f~d_:=If[DateDifference[a=DateObject[DatePlus[d[[1]],Solve[t 365.25==Tr@(x-Join[{0},
QuantityMagnitude@(DateDifference[d[[1]],#,&quot;Day&quot;]&amp;/@Rest@d)]),{x}][[1,1,2]]]/.
{y_,m_,n_,h_}:&gt;{y,m,n+1}],d[[-1]]]&gt;Quantity[0,&quot;Days&quot;],f[t,Most@d],a]


Test Cases
f[10, {{2000, 1, 1}}]
f[100, {{1975, 3, 5}, {1978, 7, 23}, {2008, 11, 12}, {2012, 12, 20}}]
f[100, {{1975, 3, 6}, {1978, 7, 23}, {2008, 11, 12}, {2012, 12, 20}}]
f[100, {{1975, 3, 9}, {1978, 7, 23}, {2008, 11, 12}, {2012, 12, 20}}]
f[100, {{1975, 3, 10}, {1978, 7, 23}, {2008, 11, 12}, {2012, 12,  20}}]
f[3, {{1990, 8, 1}, {1995, 7, 2}}]
f[50, {{1975, 3, 5}, {1978, 7, 23}, {2008, 11, 12}, {2012, 12, 20}}]