On a conditional command: more powerful than \ifnum

In order to reproduce the picture, the test should be 1 ≤ x < y.

You can parametrize the rotation angle and test for the two special cases.

\documentclass{standalone}
\usepackage{tikz}

\begin{document}
\begin{tikzpicture}
\foreach \x in {1,...,4} {
  \foreach \y in {1,...,4} {
    \def\rotation{-45}
    \ifnum\y=1
      \ifnum\x=1 \def\rotation{0} \fi
      \ifnum\x=2 \def\rotation{0} \fi
    \fi
    \fill[red!50] (\x,\y) ellipse [x radius=3pt, y radius=6pt, rotate=\rotation];
    \ifnum \x<\y \unless\ifnum \x<1 \breakforeach \fi\fi
  }
}
\draw [|-|] (.895,1) -- ++(0.211,0);
\end{tikzpicture}
\end{document}

You can also use \ifthenelse, of course.

\documentclass{standalone}
\usepackage{tikz}
\usepackage{xifthen}

\begin{document}
\begin{tikzpicture}
\foreach \x in {1,...,4} {
  \foreach \y in {1,...,4} {
    \ifthenelse{\y=1 \AND \(\x=1 \OR \x=2\)}{\def\rotation{0}}{\def\rotation{-45}}
    \fill[red!50] (\x,\y) ellipse [x radius=3pt, y radius=6pt, rotate=\rotation];
    \ifthenelse{ \x<\y \AND \NOT\(\x<1\) }{\breakforeach}{}
  }
}
\draw [|-|] (.895,1) -- ++(0.211,0);
\end{tikzpicture}
\end{document}

With a somewhat easier syntax, see https://tex.stackexchange.com/a/467527/4427

\documentclass[border=4]{standalone}
\usepackage{tikz}
\usepackage{xparse}

\ExplSyntaxOn
\NewExpandableDocumentCommand{\xifthenelse}{mmm}
 {
  \bool_if:nTF { #1 } { #2 } { #3 }
 }

\cs_new_eq:NN \numtest     \int_compare_p:n
\cs_new_eq:NN \oddtest     \int_if_odd_p:n
\cs_new_eq:NN \fptest      \fp_compare_p:n
\cs_new_eq:NN \dimtest     \dim_compare_p:n
\cs_new_eq:NN \deftest     \cs_if_exist_p:N
\cs_new_eq:NN \namedeftest \cs_if_exist_p:c
\cs_new_eq:NN \eqdeftest   \token_if_eq_meaning_p:NN
\cs_new_eq:NN \streqtest   \str_if_eq_p:ee
\cs_new_eq:NN \emptytest   \tl_if_blank_p:n
\prg_new_conditional:Nnn \xxifthen_legacy_conditional:n { p,T,F,TF }
 {
  \use:c { if#1 } \prg_return_true: \else: \prg_return_false: \fi:
 }
\cs_new_eq:NN \boolean \xxifthen_legacy_conditional_p:n
\ExplSyntaxOff

\begin{document}
\begin{tikzpicture}
\foreach \x in {1,...,4} {
  \foreach \y in {1,...,4} {
    \xifthenelse{\numtest{\y=1} && (\numtest{\x=1} || \numtest{\x=2})}
      {\def\rotation{0}}
      {\def\rotation{-45}}
    \fill[red!50] (\x,\y) ellipse [x radius=3pt, y radius=6pt, rotate=\rotation];
    \xifthenelse{ \numtest{1<=\x<\y} }{\breakforeach}{}
  }
}
\draw [|-|] (.895,1) -- ++(0.211,0);
\end{tikzpicture}
\end{document}

LaTeX's ifthen package has some facility for combining conditionals with \and and \or, \not and parentheses. But your case is easy to do with \ifnum:

\ifnum \x<\y \ifnum \x>1
  \breakforeach
\fi\fi