= Solution
We take $\mathbb N=\{1,2,\ldots\}$. An <ultrafilter> $p$ on $\mathbb N$ is a proper <filter on a set> which decides every subset: it contains $\mathbb N$ but not $\varnothing$, is closed under finite <intersections> and passage to supersets, and contains exactly one of $A$ and $\mathbb N\setminus A$ for every $A\subseteq\mathbb N$. A <principal ultrafilter> has the form $\delta_n=\{A:n\in A\}$; all other <ultrafilters> are <nonprincipal ultrafilters>.
To prove existence, start with the <cofinite filter> and order its proper filter extensions by inclusion. The union of a chain is again a proper <filter on a set>: the finitely many sets involved in an intersection all belong to one member of the chain, and no member contains the empty set. Thus <Zorn's lemma> supplies a maximal proper filter $p$. If $A\notin p$, adjoining $A$ must make the generated filter improper; equivalently, some $B\in p$ has $A\cap B=\varnothing$. Then $B\subseteq\mathbb N\setminus A$, so $\mathbb N\setminus A\in p$. This proves that $p$ is an <ultrafilter>. It contains the complement of every finite set and hence no finite set. In particular, it contains no singleton and is \b[nonprincipal].
The <Stone-Čech compactification of the natural numbers> $\beta\mathbb N$ is the set of all <ultrafilters> with basic <open sets>
$$
\widehat A=\{p:A\in p\},\qquad A\subseteq\mathbb N.
$$
We have $\widehat A\cap\widehat B=\widehat{A\cap B}$ and $\beta\mathbb N\setminus\widehat A=\widehat{\mathbb N\setminus A}$, so these basic sets define a <topology> and are <clopen sets>. If $p\ne q$, choose $A\in p\setminus q$. The disjoint basic <neighbourhoods> $\widehat A$ and $\widehat{\mathbb N\setminus A}$ separate them. Thus the space is <Hausdorff>.
For <compactness>, it suffices to show every basic open cover has a finite subcover, since an arbitrary open cover can be refined by basic <open sets>. Suppose $\{\widehat{A_i}:i\in I\}$ has no finite subcover. The complements $\mathbb N\setminus A_i$ have the <finite intersection property>: if a finite intersection were empty, the corresponding finite family of $\widehat{A_i}$ would cover every <ultrafilter>. Their generated proper <filter on a set> extends to an <ultrafilter>, by the same <Zorn's lemma> argument. This <ultrafilter> contains every $\mathbb N\setminus A_i$, and therefore belongs to none of the alleged cover members, a contradiction. We have proved that $\beta\mathbb N$ is a \b[compact Hausdorff space]. The map $n\mapsto\delta_n$ embeds the discrete natural numbers densely: any nonempty $\widehat A$ contains $\delta_n$ for some $n\in A$.
For <addition on the Stone-Čech compactification of the natural numbers>, define
$$
A-n=\{m\in\mathbb N:n+m\in A\},\qquad D_q(A)=\{n:A-n\in q\},
$$
and put
$$
A\in p+q\quad\Longleftrightarrow\quad D_q(A)\in p.
$$
The map $D_q$ preserves intersections and complements, because translation does and $q$ is an <ultrafilter>. It also sends $\mathbb N$ to $\mathbb N$ and the empty set to itself. Consequently the displayed rule defines an <ultrafilter> $p+q$. On principal points it satisfies $\delta_n+\delta_m=\delta_{n+m}$, so it extends ordinary addition.
Write $(\forall^p n)\,\varphi(n)$ to mean that $\{n:\varphi(n)\}\in p$. The <ultrafilter> addition rule is then an iterated quantifier, with its order fixed. Expansion gives
$$
\begin{aligned}
A\in(p+q)+r
&\Longleftrightarrow (\forall^p n)(\forall^q m)(\forall^r \ell)\,[n+m+\ell\in A],\\
A\in p+(q+r)
&\Longleftrightarrow (\forall^p n)(\forall^q m)(\forall^r \ell)\,[n+(m+\ell)\in A].
\end{aligned}
$$
Ordinary addition is an <associative operation>, and the quantifiers are in the same order on both lines. Thus $\boxed{(p+q)+r=p+(q+r)}$. This argument does not interchange <ultrafilter> quantifiers and does not assert that this extended addition is commutative.
For fixed $q$, the translation $R_q:p\mapsto p+q$ has
$$
R_q^{-1}(\widehat A)=\widehat{D_q(A)},
$$
a basic <clopen set>. Therefore addition is \b[continuous in its left variable]. This is the left-continuity convention used here: right translations are continuous, giving a <left-topological semigroup>. Continuity in both variables jointly is not part of the assertion.
\b[Hindman's theorem] says that every <finite colouring> of the <positive integers> has an increasing sequence $x_1<x_2<\cdots$ whose <finite-sums set>
$$
\operatorname{FS}(x_1,x_2,\ldots)=\left\{\sum_{i\in F}x_i:\varnothing\ne F\subseteq\mathbb N\text{ finite}\right\}
$$
is <monochromatic>. Assume, as permitted, that the addition just defined has an <idempotent ultrafilter> $p$, so $p+p=p$. This <ultrafilter> is nonprincipal: a principal $\delta_n$ would satisfy $\delta_n+\delta_n=\delta_{2n}\ne\delta_n$ for positive $n$. Hence every member of $p$ is infinite and every <cofinite set> belongs to $p$.
Exactly one cell $A$ of the finite colour partition belongs to $p$. Define its star set by
$$
A^*=A\cap D_p(A)=\{n\in A:A-n\in p\}.
$$
Since $A\in p+p$, we have $D_p(A)\in p$ and therefore $A^*\in p$. Moreover, for each $n\in A^*$, the set $A-n$ belongs to $p$. Idempotence gives $D_p(A-n)\in p$, and translation of the defining formula shows
$$
D_p(A-n)=D_p(A)-n,\qquad
A^*-n=(A-n)\cap D_p(A-n)\in p.
$$
This proves the <idempotent-ultrafilter star-set lemma> rather than merely invoking it.
Choose $x_1\in A^*$. Suppose $x_1<\cdots<x_k$ have been chosen with every nonempty finite sum in $A^*$. Choose $x_{k+1}$ from
$$
A^*\cap\bigcap_{s\in\operatorname{FS}(x_1,\ldots,x_k)}(A^*-s)\cap\{n:n>x_k\}.
$$
This is a finite intersection of members of the <ultrafilter> $p$, and so it is nonempty. The new sums are $x_{k+1}$ and $s+x_{k+1}$; all lie in $A^*$ by the choice. <Mathematical induction> yields an increasing infinite sequence with
$$
\boxed{\operatorname{FS}(x_1,x_2,\ldots)\subseteq A^*\subseteq A.}
$$
Thus the assumed idempotent gives the full <Hindman theorem>, with distinct increasing generators and all their nonempty finite sums of one colour.
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