Prove that a subgraph $H$ of a tournament $T$ is connected 
Consider a tournament on $2021$ vertices where for each vertex its indegree and outdegree are equal. Prove that any induced subgraph on $1348$ vertices is connected.

First, it's obvious that indegree and outdegree of all vertices equal $1010$. Since there is $1348$ vertices in a subgraph the indegree and outdegree of a vertex there is at least $1010 - (2021 - 1348) = 337$.
That's as far as I can get. Do you have any idea what I can do next?
 A: In the context of this question, I'll assume "connected" means "strongly connected"

$\qquad$ https://en.wikipedia.org/wiki/Directed_graph#Directed_graph_connectivity

as defined in the above link.

Claim:$\;$If $T$ is a tournament with $|V(T)|=2021$ such that for all $x\in V(T)$ we have $\text{indeg}_T(x)=\text{outdeg}_T(x)$, then any induced subgraph $S$ of $T$ with $|V(S)|=1348$ is such that $S$ is connected.

Proof:

Assume the hypothesis, and suppose $S$ is an induced subgraph of $T$ with $|V(S)|=1348$ which is not connected.

Our goal is to derive a contradiction.

For $x\in V(S)$, let $R(x)$ be the set of all $y\in V(S)$ for which there is a path in $S$ from $x$ to $y$.

For $x\in V(S)$, call $x$ universal if $R(x)=V(S)$.

Let $A$ be the set of elements of $V(S)$ which are universal.

Let $w\in V(S)$ be such that $|R(w)|$ is maximized.

Suppose $w\not\in A$.

Let $x\in V(S){\setminus}R(w)$.

From $x\not\in R(w)$, we get $wx\not\in E(S)$, hence $xw\in E(S)$.

But then $|R(x)|=|R(w)|+1$, contrary to the maximality of $|R(w)|$.

Thus $w\in A$, hence $A\ne{\large{\varnothing}}$.

Let $B=V(S){\setminus}A$.

Since $S$ is not connected, it follows that $B\ne{\large{\varnothing}}$.

Letting $a=|A|$ and $b=|B|$ we have $a,b > 0$ and $a+b=1348$.

Since the elements of $B$ are not universal,

*

*Any edge that ends in $A$ must start in $A$.$\\[4pt]$

*Any edge that starts in $B$ must end in $B$.

It follows that

*

*There are exactly ${\large{\binom{a}{2}}}$ edges which end in $A$.$\\[4pt]$

*There are exactly ${\large{\binom{b}{2}}}$ edges which start in $B$.

As you correctly noted, for all $x\in V(S)$ we have $\text{indeg}_S(x)\ge 337$ and $\text{outdeg}_S(x)\ge 337$.

It follows that

*

*${\large{\binom{a}{2}}}\ge 337a$, which yields $a\ge 675$.$\\[4pt]$

*${\large{\binom{b}{2}}}\ge 337b$, which yields $b\ge 675$.

But then $a+b\ge 675+675=1350$, contrary to $a+b=1348$.

This completes the proof.
A: Every tournament is a weakly connected digraph; by "connected" you mean strongly connected, i.e., there is a directed path from any vertex to any other vertex.
Assume for a contradiction that $H$ is an induced subgraph of $T$ on 1348 vertices which is not strongly connected. Then the vertex set of $H$ can be partitioned into two nonempty sets $X$ and $Y$ so that there are no directed edges from $Y$ to $X$, i.e., there are directed edges from every vertex in $X$ to every vertex in $Y$. (Namely, take two vertices $x$ and $y$ such that there is no directed path from $y$ to $x$; let $Y$ be the set of all vertices of $H$ that can be reached by a directed path from $y$, and $X$ the complement.) Without loss of generality we can assume that $|X|\le|Y|$ (otherwise reverse all the edges of $T$). Let $x=|X|$ and $y=|Y|$, so that $x+y=1348$ where $x\ge1$ and $y\ge\frac{1348}2=674$.
Now consider the subtournament induced by $X$. Since the average outdegree of a vertex in that subtournament is $\frac{x-1}2$, we can choose a vertex $u\in X$ whose outdegree relative to $X$ is $\operatorname{od}_X(u)\ge\frac{x-1}2$. Since $u$ also has edges directed to all vertices in $Y$, and since $y\ge674$, we have
$$\operatorname{od}_T(u)\ge\frac{x-1}2+y=\frac{x+y}2+\frac{y-1}2=674+\frac{y-1}2\ge1010.5\gt1010,$$
a contradiction.
