Let we follow how distance $d_n$ between this persons changes. Initially it is equal to $d_0=10$. After one time interval both persons can step right or left and distance stay the same. Or they can step in different directions and move away from each other. Or they can step into each other and the distance between them will decrease.
$$
\mathbb P(d_{n+1}=k-2|d_n=k)=\frac14,\quad \mathbb P(d_{n+1}=k+2|d_n=k)=\frac14, \quad \mathbb P(d_{n+1}=k|d_n=k)=\frac12.
$$
To get $d_7 = 0$ exactly after $7$ time intervals, you need either to have the event $d_n \mapsto d_n-2$ five times and twice $d_n$ remain unchanged, or once $d_n$ is increased and $6$ times decreased. Therefore
$$
\mathbb P(d_7=0\,|\,d_0=10)=\binom{7}{2}\left(\frac14\right)^5\left(\frac12\right)^2+\binom{7}{1}\left(\frac14\right)^6\frac14 = \dfrac{91}{4^7}=\dfrac{91}{16384}.
$$
If meetings before $n=7$ should be also taken into account, one calculate probability $\mathbb P(d_5=0 \vee d_6=0 \vee d_7=0)$ directly. Mark step
"down" $d_n \mapsto d_n-2$ by $d$, step "right" $d_n \mapsto d_n$ by $r$ and step "up" $d_n \mapsto d_n+2$ by $u$.
$$\mathbb P(d_5=0 \vee d_6=0 \vee d_7=0)=\mathbb P(d_5=0)+\mathbb P(d_5\neq 0, d_6=0)+\mathbb P(d_5\neq 0, d_6\neq 0, d_7=0).$$
$$\mathbb P(d_5=0)=\mathbb P(ddddd)=\dfrac{1}{4^5},$$
$$\mathbb P(d_5\neq 0, d_6=0) = \mathbb P(\underbrace{rddddd \vee\ldots\vee ddddrd}_{5}) =5\dfrac{1}{2}\dfrac{1}{4^5}$$.
Note that the last step cannot be $r$ here. Next,
$$\mathbb P(d_5\neq 0, d_6\neq 0, d_7=0) = \mathbb P(\underbrace{udddddd\vee\ldots\vee ddddudd}_{5} \vee \underbrace{rrddddd\vee\ldots\vee ddddrrd}_{\binom62})=
$$
$$=5\dfrac{1}{4^7}+\binom62 \dfrac{1}{2^2}\dfrac{1}{4^5}.$$
Here $\binom62$ variants appears since last step should be $d$ and two letters $r$ can be at any other six places.
Total probability is
$$\mathbb P(d_5=0 \vee d_6=0 \vee d_7=0)=\dfrac{1}{4^5}+5\dfrac{1}{2}\dfrac{1}{4^5}+5\dfrac{1}{4^7}+\binom62 \dfrac{1}{2^2}\dfrac{1}{4^5}=\dfrac{121}{4^7}.$$
Let me solve a related but slightly different problem. Consider a discrete random walk in 2D with $1/4$ probability of moving in each of the four directions for each step. We are given a destination vector $\vec{D}=Xe_1+Ye_2$ and we start at the origin. The question is: Given $n$, what is the probability $P_n$ of ending up at $X$ after $n$ steps.
By an $n$-path, I mean a sequence $\vec{r}_1, \cdots, \vec{r}_n$ with each $\vec{r}=xe_1+ye_2$ being such that either $x=\pm 1$ and $y=0$ or $x=0$ and $y=\pm 1$. There are exactly $4^n$ different $n$-paths. Let $N_n(X,Y)$ be the number of $n$-paths that end at $\vec{D}$. Then $P_n=N_n(X,Y)/4^n$. So we need to find $N_n(X,Y)$.
Take an $n$-path. Let $N^x_{\pm}$ be the number of horizontal steps in $\pm$ directionrespectively. Similarly define $N_\pm^y$. Clearly,
$$
N_+^x-N_-^x=X, \qquad N_+^y-N_-^y=Y, \qquad N_+^x+N_-^x+N_+^y+N_-^y=n
$$
We furthermore say an $n$-path is $m$-horizontal if $N_+^x+N_-^y=m$. Forgetting for the moment that we only care about non-negative integer solutions, one can solve these four equations
$$
\begin{pmatrix}
N_+^x\\
N_-^x\\
N_+^y\\
N_-^y
\end{pmatrix}=\frac{1}{2}\begin{pmatrix}
m+X\\
m-X\\
n-m+Y\\
n-m-Y
\end{pmatrix}
$$
Which obiously requires $m\geq |X|$, $n-m\geq |Y|$ and that $X+m$ and $n-m+Y$ are even. We say $m$ is $(n,X,Y)$-admissible if given $n,X,Y$, the number $m$ is such that all these conditions are satisfied. Then the number of $m$-horizontal $n$-paths ending at $\vec{D}$ are
$$
\nu_m=\binom{n}{m}\binom{m}{N_+^x}\binom{n-m}{N_+^y}=
\binom{n}{m}\binom{m}{\frac{m+X}{2}}\binom{n-m}{\frac{n-m+Y}{2}}
$$
and the total number of $n$-paths ending at $\vec{D}$ are
$$
\boxed{
N_n(X,Y)=\sum_{m\text{ is }(n,X,Y)\text{-admissible}}
\binom{n}{m}\binom{m}{\frac{m+X}{2}}\binom{n-m}{\frac{n-m+Y}{2}}}
$$
Note that it is not hard at all to find all $m$ that are $(n,X,Y)$-admissible once you know specifically what $n,X,Y$ are. For example, say $X=-10, Y=30$ and $n=1000$. Since $X$ is even, $m$ has to be even. The condition $n-m+Y$ even is automatically satisfied then. So you need $m\geq 10$ and $1000-m\geq 30$. So the set of $(1000, -10, 30)$-admissible $m$'s are all even numbers satisfiying $10\leq m\leq 970$.
Best Answer
Here are some easy remarks on the question (but not a solution). First, if $n$ is odd the two particles never meet, for parity reasons, hence one can assume that $n$ is even.
Say that the particle starting from $0$ hits $n$ at time $T_n$. The particles cannot cross each other without meeting hence, at time $T_n$, they already met. Thus, the meeting time $M_n$ is almost surely finite and $$E(M_n)\leqslant E(T_n)=n^2. $$ Note that $M_n$ is also at most the hitting time $T'_n$ of $0$ by the particle starting from $n$, hence $$E(M_n)\leqslant E(\min(T_n,T'_n)),$$ where $T'_n$ is an independent copy of $T_n$. Conversely, the particles cannot meet before at least one of them hits $n/2$ hence $$E(M_n)\geqslant E(\min(T_{n/2},T'_{n/2})). $$ This suggests that there might exist some finite positive $\mu$ such that $E(M_n)\sim\mu n^2$ when $n\to\infty$. If this is so, one knows that $\frac14\theta\leqslant\mu\leqslant\theta$ where $\theta$ in $(0,1)$ is defined by the equivalent $E(\min(T_n,T'_n))\sim\theta n^2$.
Note finally that $\min(T_n,T'_n)$ is the first hitting time of the boundary of the square $[-n,n]^2$ by a random walk on $\mathbb Z^2$ starting from $(0,0)$ with steps $(\pm1,\pm1)$.