*(I am late to post a very similar answer to the already given one, yet I'd like to post it as well, since it differs in some detail)*

Changing the summation index to $m=n+\ell$, the identity writes
$$\sum_{m=\ell}^{R-t+\ell } {m\choose \ell}{R-m\choose t-\ell}={R+1\choose t+1}.$$

Given natural numbers $\ell\le t\le R$, and $m$, we may consider the class of those $(t+1)$-subsets $\{x_0<x_1<\dots<x_t\}$ of $\{0,1,\dots,R\}$ such that $x_\ell=m$: these are exactly ${m\choose \ell}{R-m\choose t-\ell}$ (indeed the $\ell$ elements $x_0,\dots, x_{\ell-1}$ can be chosen freely into $\{0,\dots, m-1\}$, and so can the $t-\ell$ elements $x_{\ell+1},\dots,x_t$ into $\{m+1,\dots,R\}$. These classes, for $ \ell\le m\le R-t+\ell $ form a partition of all $(t+1)$-subsets of $[R+1]$, whence the sum of their cardinality is independent of $\ell$ and the identity.