= Solution
The coefficient of $x^{n+1}$ in $r_j$ is
$$
c_j=\frac1{\prod_{k\ne j}(y_j-y_k)}.
$$
With the increasing enumeration, its sign is $(-1)^{n+2-j}$. Therefore all terms $(-1)^jc_j$ have the same sign $(-1)^{n+2}$, and $R:=\sum_j(-1)^jc_j\ne0$. The leading coefficient of $t$ is $T:=\sum_jf(y_j)c_j$. Canceling the coefficient of $x^{n+1}$ gives the unique value
$$
\boxed{\lambda=\frac TR=\frac{\sum_j f(y_j)/\prod_{k\ne j}(y_j-y_k)}{\sum_j(-1)^j/\prod_{k\ne j}(y_j-y_k)}.}
$$
Both interpolants have degree at most $n+1$, so this single cancellation is equivalent to $\deg(t-\lambda r)\leq n$.
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