Solution:
We claim that if $2026$ is replaced with $2n$, the answer is $\frac{2n}{n + 1}$.

Write the path as a sequence of $U$, $D$, $L$, and $R$ moves. The possible sequences that can result are precisely those with an equal number of $U$'s and $D$'s, and an equal number of $R$'s and $L$'s. We first project this sequence onto a single dimension by converting each $U$ and $R$ to a $1$, and each $D$ and $L$ to a $-1$. The resulting sequence will have an equal number of $1$'s and $-1$'s.

We claim that every such sequence of $n$ $1$'s and $n$ $-1$'s corresponds to the same number of closed lattice loops. Indeed, given such a sequence, a corresponding lattice loop can be made by replacing all $1$'s with $U$'s and $R$'s, and all $-1$'s with $D$'s and $L$'s, so that there are an equal number of $U$'s and $D$'s. Nothing in this replacement process is order-dependent, so the number of ways to create this loop does not depend on the initial sequence.

We can see that each of the $1$'s move the path towards $\ell$ and the $-1$'s move the path away. Hence, we can think of the path as a one-dimensional walk, starting and ending in the same place, where the points on $\ell$ correspond exactly to the maxima of this one-dimensional walk.

Using Catalan numbers, the probability that any given point is at the maxima is
$$\frac{\frac{1}{n + 1}\binom{2n}{n}}{\binom{2n}{n}} = \frac{1}{n + 1},$$
and thus by linearity, the expected number of $i \in [0, 2025]$ such that $P_{i}$ is on $\ell$ is $\frac{2n}{n + 1}$.

Plugging in $n = 1013$ gives a final answer of $\left[\frac{1013}{507}\right]$.