The bijective argument for all $p$ is the following. Write $n = ap + b$ where $0 \le b \le p-1$, so that $a = \lfloor \frac{n}{p} \rfloor$. Divide the set $[n] = \{ 1, 2, \dots n \}$ into $a$ groups of $p$ elements and $b$ elements left over. Consider the action of the cyclic group $C_p$ on the set of $p$-element subsets of $n$ by cyclic permutation on each of the $a$ groups of $p$ elements. There are two kinds of orbits, orbits of size $p$ and fixed points, so ${n \choose p}$ is congruent $\bmod p$ to the number of fixed points. And the fixed points are exactly given by the $a$ groups of $p$ elements themselves, of which there are $a = \lfloor \frac{n}{p} \rfloor$.
A generalization of this argument proves that
$${ap + b \choose cp + d} \equiv {a \choose c} {b \choose d} \bmod p$$
and iterating this identity proves Lucas' theorem
$${\sum a_i p^i \choose \sum b_i p^i} \equiv \prod {a_i \choose b_i} \bmod p$$
where $a_i, b_i$ are digits in base $p$; this can also be proven directly with a similar argument. You can see several other arguments like this at this blog post, including a bijective proof of Fermat's little theorem and Wilson's theorem.
An important corollary of this result is that if $p^k$ is the largest power of $p$ dividing $n$ then ${n \choose p^k}$ is not divisible by $p$ (which also follows from Kummer's theorem). This fact can famously be used to prove the first Sylow theorem.
Edit: Stripping out the group theory, here is the argument specialized to the case $p = 3$ for concreteness but there's nothing special about $3$ here. Write $n = 3a + b$ where $0 \le b \le 2$. Divide the set $[n] = \{ 1, 2, \dots 3a + b \}$ into $a$ groups of $3$ elements
$$\{ 1, 2, 3 \}, \{ 4, 5, 6\}, \dots \{3a-2, 3a-1, 3a \}$$
together with $b$ leftover elements $\{ 3a+1, \dots 3a+b \}$. Now we are going to group together the $3$-element subsets of $\{ 1, 2, \dots 3a+b \}$ as follows:
- There are $a$ special $3$-element subsets given by the groups $\{ 1, 2, 3 \}, \{ 4, 5, 6 \}$, etc. we just picked.
- All of the other $3$-element subsets can be organized into groups of $3$ as follows. Consider the function $f : [n] \to [n]$ which "rotates" each of the $3$-element sets by adding $1 \bmod 3$ to each of them; that is, $f(1) = 2, f(2) = 3, f(3) = 1, f(4) = 5, f(5) = 6, f(6) = 4$, etc. $f$ does not do anything to the "remainder" $\{ 3a+1, \dots 3a+b \}$. Then every $3$-element subset $\{ i, j, k \}$ not of the above form is matched up with exactly two other $3$-element subsets $\{ f(i), f(j), f(k) \}, \{ f(f(i)), f(f(j)), f(f(k)) \}$ under the action of $f$. For example, $\{ 1, 2, 4 \}$ is matched up with $\{ 2, 3, 5 \}$ and $\{ 3, 1, 6 \}$. The $a$ special $3$-element subsets are exactly the subsets with the property that $\{ i, j, k \} = \{ f(i), f(j), f(k) \}$, so they don't get matched up with anything by $f$.
The general result, again stripped of any explicit references to group theory, is the following. Suppose $p$ is a prime, $X$ is a finite set, and $f : X \to X$ is a permutation such that $f^p(x) = x$ for all $x \in X$. Then $X$ splits up as the disjoint union of the fixed points of $f$ together with subsets of size $p$ of the form $\{ x, f(x), f^2(x), \dots f^{p-1}(x) \}$; in particular, $|X|$ is congruent to the number of fixed points of $f$, $\bmod p$.
Let $g_n(x)$ be the generating function of row $n$ of your triangle:
$g_0(x) = 1$, $g_1(x) = 1+x$, $g_2(x) = 1 + x^2$, etc.
It seems [empirically: I don't have a proof] these satisfy the recurrence
$$4 x (1+x) n g_n(x) - 4 x n g_{n + 1}(x) - (n+3)(1+x) g_{n+2}(x) + (n + 3) g_{n+3}(x) = 0 $$
The sum of row $n$ is then $g_n(1)$, which satisfies
$$ 8 n g_n(1) - 4 n g_{n+1}(1) - 2 (n+3) g_{n+2}(1) + (n+3) g_{n+3}(1) = 0 $$
and the solution of this with initial conditions $g_0(1) = 1$, $g_1(1) = 2$, $g_2(1) = 2$ is
$$ g_{2n}(1) = {2n \choose n},\ g_{2n+1}(1) = 2 {2n \choose n} $$
EDIT: OK, here's a proof. Let $h_k(x)$ be $g_k(x)$ with the signs of the coefficients of $x^i$ for $i > k/2$ changed. Thus
$$ \eqalign{h_1(x) &= 1-x \cr
h_2(x) &= 1 - x^2\cr
h_3(x) &= 1 + x - x^2 - x^3\cr
h_4(x) &= 1 + 2 x - 2 x^3 - x^4\cr} $$
The coefficient of $x^{k}$ in $h_n(x)$ is $-$ the coefficient of $x^{n-k}$ there. It can then be seen that the coefficients of $h_n$ form the entries of a "Pascal's triangle" starting with the row $(1,-1)$; there's no need to kill the central coefficients because the symmetry makes them automatically $0$. Thus $h_n(x) = (1-x) (1+x)^{n-1}$. So the coefficient of $x^k$ in $g_n(x)$ is ${n-1 \choose k} - {n-1 \choose k-1}$ for $k < n/2$ (with ${n-1 \choose -1} = 0$) and
${n-1 \choose k-1} - {n-1 \choose k}$ for $k > n/2$. Adding these up,
taking advantage of telescoping, we have
$$ \eqalign{g_{2n}(1) &= 2 \sum_{k=0}^{n-1} \left({2n-1 \choose k} - {2n-1 \choose k-1}\right) = 2{2n-1 \choose n-1} = {2n \choose n}\cr
g_{2n+1}(1) &= 2 \sum_{k=0}^{n} \left({2n \choose k} - {2n \choose k-1}\right) = 2{2n \choose n}\cr}$$
Best Answer
Preliminaries
We will use the following
Theorem (Jitsuro Nagura). There exists at least one prime number $p$ such as $$ a_m\leq n < p < \left(1+\frac{1}{m}\right)n\tag1 $$ where $m=1,2,3,4,5$ and $a_m=2,8,9,24,25$, respectively.
Proof. See "On The Interval Containing At Least One Prime Number" by Jitsuro Nagura in Proc. Japan Acad. 28(4): 177-181 (1952). DOI: 10.3792/pja/1195570997. Accessible for example here.$\square$
Proof of conjecture
Theorem. Let $n$ be a positive integer. If $$ P_n=\prod\limits_{k=0}^n\frac{(2n)!k!}{(2k)!(2n-k)!}=\prod\limits_{k=1}^n\frac{(2n)!k!}{(2k)!(2n-k)!}\tag2 $$ is an integer, then $n\in\{1,2,5\}$.
Proof. Calculation shows that if $1\leq n\leq 8$, then $P_n$ is an integer only for $n\in\{1,2,5\}$. Assume $n\geq 9$. By Nagura's theorem, there exists at least one prime $p$ with $n<p<\left(1+\frac{1}{3}\right)n$. Denote by $k_p(a)$ the number of times prime $p$ appears in the prime factorization of an integer $a$ and define $$ \begin{align} A_n&:=\prod_{k=1}^n(2n)!k!\tag3\\ B_n&:=\prod_{k=1}^n(2k)!\tag4\\ C_n&:=\prod_{k=1}^n(2n-k)!\tag5 \end{align} $$ so that $P_n=\frac{A_n}{B_nC_n}$. Now, $n<p<2n$, so $k_p(A_n)=n$. Moreover, $p<\frac{4n}{3}$, so $k_p(B_n)>\frac{n}{3}$ and $k_p(C_n)>\frac{2n}{3}$. Therefore, $k_p(B_nC_n)>n=k_p(A_n)$ and we conclude that if $n\geq 9$, then $P_n\notin\mathbb{Z}$.$\square$