Day of the year

You can use PGF's calendar library to convert the current day and the first day of the current year into julian dates:

\documentclass{article}

\usepackage{pgfkeys, pgfcalendar}

\newcount\julianA
\newcount\julianB

\newcommand\doy{%
    \pgfcalendardatetojulian{\year-\month-\day}{\julianA}%
    \pgfcalendardatetojulian{\year-1-1}{\julianB}%
    \advance\julianA by -\julianB%
    \advance\julianA by 1\relax%
    \the\year\the\julianA%
}

\begin{document}
Day of the year: \doy
\end{document}

My attempt with the powerful expl3. I know the interface is not good at all, but please bear with me, I'm stupid a newbie. :)

\documentclass{article}

\usepackage{expl3}
\usepackage{xparse}

\ExplSyntaxOn

\bool_new:N \l_leap_year_bool
\int_new:N \l_day_of_year_int
\cs_generate_variant:Nn \int_to_arabic:n { V }

\DeclareDocumentCommand \dayoftheyear { m m m }
{
  \bool_set:Nn \l_leap_year_bool
  {
    \int_compare_p:n { \int_mod:nn { #3 } { 4 } = 0 } &&
    \int_compare_p:n { \int_mod:nn { #3  } { 100 } != 0 } ||
    \int_compare_p:n { \int_mod:nn { #3  } { 400 } = 0 }
  }
  \int_case:nnn { #2 - 1 }
  {
    { 1 } { \int_set:Nn \l_day_of_year_int { 31 } }
    { 2 } { \int_set:Nn \l_day_of_year_int { 59 } }
    { 3 } { \int_set:Nn \l_day_of_year_int { 90 } }
    { 4 } { \int_set:Nn \l_day_of_year_int { 120 } }
    { 5 } { \int_set:Nn \l_day_of_year_int { 151 } }
    { 6 } { \int_set:Nn \l_day_of_year_int { 181 } }
    { 7 } { \int_set:Nn \l_day_of_year_int { 212 } }
    { 8 } { \int_set:Nn \l_day_of_year_int { 243 } }
    { 9 } { \int_set:Nn \l_day_of_year_int { 273 } }
    { 10 } { \int_set:Nn \l_day_of_year_int { 304 } }
    { 11 } { \int_set:Nn \l_day_of_year_int { 334 } }
  }
  { }
  \bool_if:nT 
  {
    \l_leap_year_bool && \int_compare_p:n { #2 > 2}
  } { \int_incr:N \l_day_of_year_int }
  \int_add:Nn \l_day_of_year_int { #1 }
  \int_to_arabic:V \l_day_of_year_int
}
\ExplSyntaxOff

\begin{document}

Day of the year: \dayoftheyear{27}{6}{2013}

\end{document}

Hope it helps. :)

I knew that one day my \juliandate_calc:nnnn macro would have been useful. ;-)

\documentclass{article}

\usepackage{xparse}

\ExplSyntaxOn

\NewDocumentCommand{\dayofyear}{o}
 {
  \IfNoValueTF{#1}
   {
    \aaki_day_of_year:nnn { \year } { \month } { \day }
   }
   {
    \aaki_day_of_year:n { #1 }
   }
 }
\cs_new:Npn \aaki_day_of_year:n #1
 {
  \aaki_day_of_year:w #1 \q_stop
 }
\cs_new:Npn \aaki_day_of_year:w #1 - #2 - #3 \q_stop
 {
  \aaki_day_of_year:nnn { #1 } { #2 } { #3 }
 }

\cs_new:Npn \aaki_day_of_year:nnn #1 #2 #3
 {
  \int_eval:n { #1 }
  \int_eval:n 
   {
    \juliandate_calc:nnnn { #3 } { #2 } { #1 } { \use:n }
    -
    \juliandate_calc:nnnn { 1 } { 1 } { #1 } { \use:n }
    + 1
   }
 }

\cs_new:Npn \juliandate_calc:nnnn #1 #2 #3 #4 % #1 = day, #2 = month, #3 = year, #4 = what to do
 {
  #4
   {
    \int_eval:n
     {
      #1 +
      \int_div_truncate:nn
       {
        153 * (#2 + 12 * \int_div_truncate:nn { 14 - #2 } { 12 } - 3) + 2
       }
       { 5 } +
      365 * (#3 + 4800 - \int_div_truncate:nn { 14 - #2 } { 12 } ) +
      \int_div_truncate:nn { #3 + 4800 - \int_div_truncate:nn { 14 - #2 } { 12 } } { 4 } -
      \int_div_truncate:nn { #3 + 4800 - \int_div_truncate:nn { 14 - #2 } { 12 } } { 100 } +
      \int_div_truncate:nn { #3 + 4800 - \int_div_truncate:nn { 14 - #2 } { 12 } } { 400 } -
      32045
     }
   }
 }

\ExplSyntaxOff

\begin{document}

\dayofyear

\dayofyear[2013-1-1]

\dayofyear[2013-12-31]

\dayofyear[2012-12-31]

\end{document}

The algorithm for getting the Julian date from day, month and year can be found on the net.

The internal command \aaki_day_of_year:n (for the date in ISO format YYYY-MM-DD) is fully expandable.