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Here is an argument that resticts the structure of such groups of even order, which expands on ARupinksi's comments above (inspiration came from a nice argument in this paper).
Namely, suppose that $1\neq t$ is a real element in $G$, where $G$ is as in the question. Form the following normal subgroups: $$ K = \langle t \rangle^G = \bigcap_{\chi(1)=\chi(t)} \ker \chi \quad \text{and} \quad L = \bigcap_{\chi(t) \neq 0} \ker \chi < K , $$ where the intersections run over irreducible characters $\chi$ of $G$. Then we have $$ t^G = K \setminus L,$$ that is, the elements in $K\setminus L$ are all conjugate in $G$ to $t$. In particular, it turns out that $t$ is rational in the sense that it is conjugate to every generator of $\langle t \rangle$, which is equivalent to every character having rational value at $t$.
Why is this true? Well, column orthogonality yields $\DeclareMathOperator{\Irr}{Irr}$ $$ 0 = \sum_{\chi \in \Irr G} \chi(t)\chi(1) = |G:K| + \sum_{\chi(t)<\chi(1)}\chi(t)\chi(1). $$ Now let $y\in K$ and suppose that $t$ and $y$ are not conjugate. Then plugging in $y$ instead of $1$ in the last formula, we see that $$ \sum_{\chi(t)<\chi(1)} (-\chi(t))\chi(y) = |G:K| =\sum_{\chi(t)<\chi(1)} (-\chi(t))\chi(1) .$$ Since $-\chi(t)>0$ and $|\chi(y)|\leq\chi(1)$, it follows that $y$ is in the kernel of every irr char $\chi$ with $\chi(t)< 0$, that is, $y\in L$. Thus $K\setminus L = t^G$.
(Added later:) If $t$ is an involution, then it follows that $x^t=x^{-1}$ for all $x\in L$, in particular $L$ must be abelian, and elements of $L$ are real.

The "dual" argument (exchanging the roles of characters and conjugacy classes) shows the following: Suppose $1\neq \chi$ is real valued, and let $V= \operatorname{\mathbf{V}}(\chi)$ be the vanishing-off group of $\chi$, generated by all group elements on which $\chi$ is non-zero. Then $$ \Irr( G/\ker \chi ) = \{ \chi \} \cup \Irr(G/V). $$ From this it follows easily that $V/\ker\chi$ is a conjugacy class of $G/\ker\chi$ and that the only value of $\chi$ besides $0$ and $\chi(1)$ is $-\chi(1)/(|V/\ker\chi|-1)$. (In particular, any real character is rational.) Groups with such an character have been studied by Zhmud, where more information can be found. I
I suppose there is also literature on groups having normal subgroups $L\subset K$ such that $K\setminus L$ is a conjugacy class of $G$. The notion of a Camina pair/group seems to be related (see this paper and papers that refer to it).

Here is an argument that resticts the structure of such groups of even order, which expands on ARupinksi's comments above (inspiration came from a nice argument in this paper).
Namely, suppose that $1\neq t$ is a real element in $G$, where $G$ is as in the question. Form the following normal subgroups: $$ K = \langle t \rangle^G = \bigcap_{\chi(1)=\chi(t)} \ker \chi \quad \text{and} \quad L = \bigcap_{\chi(t) \neq 0} \ker \chi < K , $$ where the intersections run over irreducible characters $\chi$ of $G$. Then we have $$ t^G = K \setminus L,$$ that is, the elements in $K\setminus L$ are all conjugate in $G$ to $t$. In particular, it turns out that $t$ is rational in the sense that it is conjugate to every generator of $\langle t \rangle$, which is equivalent to every character having rational value at $t$.
Why is this true? Well, column orthogonality yields $\DeclareMathOperator{\Irr}{Irr}$ $$ 0 = \sum_{\chi \in \Irr G} \chi(t)\chi(1) = |G:K| + \sum_{\chi(t)<\chi(1)}\chi(t)\chi(1). $$ Now let $y\in K$ and suppose that $t$ and $y$ are not conjugate. Then plugging in $y$ instead of $1$ in the last formula, we see that $$ \sum_{\chi(t)<\chi(1)} (-\chi(t))\chi(y) = |G:K| =\sum_{\chi(t)<\chi(1)} (-\chi(t))\chi(1) .$$ Since $-\chi(t)>0$ and $|\chi(y)|\leq\chi(1)$, it follows that $y$ is in the kernel of every irr char $\chi$ with $\chi(t)< 0$, that is, $y\in L$. Thus $K\setminus L = t^G$.

The "dual" argument (exchanging the roles of characters and conjugacy classes) shows the following: Suppose $1\neq \chi$ is real valued, and let $V= \operatorname{\mathbf{V}}(\chi)$ be the vanishing-off group of $\chi$, generated by all group elements on which $\chi$ is non-zero. Then $$ \Irr( G/\ker \chi ) = \{ \chi \} \cup \Irr(G/V). $$ From this it follows easily that $V/\ker\chi$ is a conjugacy class of $G/\ker\chi$ and that the only value of $\chi$ besides $0$ and $\chi(1)$ is $-\chi(1)/(|V/\ker\chi|-1)$. (In particular, any real character is rational.) Groups with such an character have been studied by Zhmud, where more information can be found. I suppose there is also literature on groups having normal subgroups $L\subset K$ such that $K\setminus L$ is a conjugacy class of $G$. The notion of a Camina pair/group seems to be related (see this paper and papers that refer to it).

Here is an argument that resticts the structure of such groups of even order, which expands on ARupinksi's comments above (inspiration came from a nice argument in this paper).
Namely, suppose that $1\neq t$ is a real element in $G$, where $G$ is as in the question. Form the following normal subgroups: $$ K = \langle t \rangle^G = \bigcap_{\chi(1)=\chi(t)} \ker \chi \quad \text{and} \quad L = \bigcap_{\chi(t) \neq 0} \ker \chi < K , $$ where the intersections run over irreducible characters $\chi$ of $G$. Then we have $$ t^G = K \setminus L,$$ that is, the elements in $K\setminus L$ are all conjugate in $G$ to $t$. In particular, it turns out that $t$ is rational in the sense that it is conjugate to every generator of $\langle t \rangle$, which is equivalent to every character having rational value at $t$.
Why is this true? Well, column orthogonality yields $\DeclareMathOperator{\Irr}{Irr}$ $$ 0 = \sum_{\chi \in \Irr G} \chi(t)\chi(1) = |G:K| + \sum_{\chi(t)<\chi(1)}\chi(t)\chi(1). $$ Now let $y\in K$ and suppose that $t$ and $y$ are not conjugate. Then plugging in $y$ instead of $1$ in the last formula, we see that $$ \sum_{\chi(t)<\chi(1)} (-\chi(t))\chi(y) = |G:K| =\sum_{\chi(t)<\chi(1)} (-\chi(t))\chi(1) .$$ Since $-\chi(t)>0$ and $|\chi(y)|\leq\chi(1)$, it follows that $y$ is in the kernel of every irr char $\chi$ with $\chi(t)< 0$, that is, $y\in L$. Thus $K\setminus L = t^G$.
(Added later:) If $t$ is an involution, then it follows that $x^t=x^{-1}$ for all $x\in L$, in particular $L$ must be abelian, and elements of $L$ are real.

The "dual" argument (exchanging the roles of characters and conjugacy classes) shows the following: Suppose $1\neq \chi$ is real valued, and let $V= \operatorname{\mathbf{V}}(\chi)$ be the vanishing-off group of $\chi$, generated by all group elements on which $\chi$ is non-zero. Then $$ \Irr( G/\ker \chi ) = \{ \chi \} \cup \Irr(G/V). $$ From this it follows easily that $V/\ker\chi$ is a conjugacy class of $G/\ker\chi$ and that the only value of $\chi$ besides $0$ and $\chi(1)$ is $-\chi(1)/(|V/\ker\chi|-1)$. (In particular, any real character is rational.) Groups with such an character have been studied by Zhmud, where more information can be found.
I suppose there is also literature on groups having normal subgroups $L\subset K$ such that $K\setminus L$ is a conjugacy class of $G$. The notion of a Camina pair/group seems to be related (see this paper and papers that refer to it).

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Here is an argument that resticts the structure of such groups of even order, which expands on ARupinksi's comments above (inspiration came from a nice argument in this paper).
Namely, suppose that $1\neq t$ is a real element in $G$, where $G$ is as in the question. Form the following normal subgroups: $$ K = \langle t \rangle^G = \bigcap_{\chi(1)=\chi(t)} \ker \chi \quad \text{and} \quad L = \bigcap_{\chi(t) \neq 0} \ker \chi < K , $$ where the intersections run over irreducible characters $\chi$ of $G$. Then we have $$ t^G = K \setminus L,$$ that is, the elements in $K\setminus L$ are all conjugate in $G$ to $t$. In particular, it turns out that $t$ is rational in the sense that it is conjugate to every generator of $\langle t \rangle$, which is equivalent to every character having rational value at $t$.
Why is this true? Well, column orthogonality yields $\DeclareMathOperator{\Irr}{Irr}$ $$ 0 = \sum_{\chi \in \Irr G} \chi(t)\chi(1) = |G:K| + \sum_{\chi(t)<\chi(1)}\chi(t)\chi(1). $$ Now let $y\in K$ and suppose that $t$ and $y$ are not conjugate. Then plugging in $y$ instead of $1$ in the last formula, we see that $$ \sum_{\chi(t)<\chi(1)} (-\chi(t))\chi(y) = |G:K| =\sum_{\chi(t)<\chi(1)} (-\chi(t))\chi(1) .$$ Since $-\chi(t)>0$ and $|\chi(y)|\leq\chi(1)$, it follows that $y$ is in the kernel of every irr char $\chi$ with $\chi(t)< 0$, that is, $y\in L$. Thus $K\setminus L = t^G$.

The "dual" argument (exchanging the roles of characters and conjugacy classes) shows the following: Suppose $1\neq \chi$ is real valued, and let $V= \operatorname{\mathbf{V}}(\chi)$ be the vanishing-off group of $\chi$, generated by all group elements on which $\chi$ is non-zero. Then $$ \Irr( G/\ker \chi ) = \{ \chi \} \cup \Irr(G/V). $$ From this it follows easily that $V/\ker\chi$ is a conjugacy class of $G/\ker\chi$ and that the only value of $\chi$ besides $0$ and $\chi(1)$ is $-\chi(1)/(|V/\ker\chi|-1)$. (In particular, any real character is rational.) Groups with such an character have been studied by Zhmud, where more information can be found. I suppose there is also literature on groups having normal subgroups $L\subset K$ such that $K\setminus L$ is a conjugacy class of $G$. The notion of a Camina pair/group seems to be related (see this paper and papers that refer to it).