I was reading the book Painleve transcendents: The Riemann-Hilbert approach and it said the following:
Suppose that the rational matrix function $A(\lambda)$ has only simple poles. Then all singular points of equation $$ \frac{d\Phi(\lambda)}{d\lambda} = A(\lambda)\Phi(\lambda) $$ are Fuchsian, and the equation itself is called Fuchsian equation. Assuming that $a_1,\dots,a_m$ are all the singular points and $a_m = \infty$, then a Fuchsian equation with $m$ regular singular points can be written as $$ \frac{d\Phi(\lambda)}{d\lambda} =\sum_{k=1}^{m-1} \frac{A_k}{\lambda-a_k}\Phi(\lambda) $$
The book then said that
Observe that if $\sum A_k = 0$, then the infinity is actually a regular point and the equation has in fact $m-1$ singular points located on the affline part of $\mathbb{C}P^1$, the one-dimensional complex projective space(isomorphic to the Riemann sphere $\overline{\mathbb{C}}$).
and I am not sure how to observe that.
To study the ODE you described near $\lambda=\infty$, it is instructive to change variables to $t=1/\lambda$ and consider the result near $t=0$.
Writing $\Phi(\lambda)=\Psi(t)$, and noticing that $$ \frac{d\Psi(t)}{dt}=-\,\lambda^2\frac{d\Phi(\lambda)}{d\lambda},$$ the ODE becomes $$ \frac{d\Psi(t)}{dt}=-\,\frac{1}{t}\sum_{k=1}^{m-1}\frac{A_k}{1-a_k\,t}\Psi(t).$$ Expand the coefficient function on the right side of this equation around $t=0$ to see that the first term in the expansion of the coefficient is $$-\,\frac{1}{t}\sum_{k=1}^{m-1}A_k.$$ So it is singular at $t=0$, unless the sum $\sum_{k=1}^{m-1}A_k$ vanishes.
If this sum does indeed vanish, then $\lambda=\infty$ is not singular, that is, the ODE is regular at infinity (in $\lambda$). (Note that infinity is not always regarded as a point, but it can be regarded as one in complex projective space $\mathbb C\mathbb P^1$. If you’re unfamiliar with such spaces, see https://en.wikipedia.org/wiki/Complex_projective_space .)