$$
\begin{aligned}
f(x)= & a_{2020} x^{2020}+a_{2019} x^{2019}+\cdots+ \\
& a_{2018} x^{2018}+a_{1} x+a_{0},
\end{aligned}
$$

where, $a_{i} \in \mathbf{Z}, i=0,1, \cdots, 2020$. Find the number of ordered pairs $(n, p)$ (where $p$ is a prime) that satisfy $p^{2}<n<2020$, and for any $i=0,1, \cdots, n, \mathrm{C}_{n}^{i}+f(i)$ is congruent modulo $p$.
Three, Lemma $1 \mathrm{C}_{n}^{0}, \mathrm{C}_{n}^{1}, \cdots, \mathrm{C}_{n}^{n}$ are all odd if and only if $n=2^{r}-1$.

Proof of Lemma 1: By a well-known conclusion, $\mathrm{C}_{n}^{i}$ is even if and only if $i+(n-i)=n$ causes a carry in binary.

By the problem statement, when $i=0,1, \cdots, n$, $i+(n-i)=n$ does not cause a carry in binary, thus,
$$
n=(11 \cdots 1)_{2} \text {. }
$$

Lemma 2: If $\mathrm{C}_{n}^{i}+\mathrm{C}_{n}^{j} \equiv i+j(\bmod 2)(0 \leqslant i < j \leqslant n)$, and $n > p^2$, set $n=\left(n_{k} \cdots n_{1} n_{0}\right)_{2}$.
Since $n > p^2$, thus, $k \geqslant 2$.
Take $i_{1}=n_{0}, i_{2}=p+n_{0}, i_{3}=2 p+n_{0}$. Then $f\left(i_{1}\right) \equiv f\left(i_{2}\right) \equiv f\left(i_{3}\right)(\bmod p)$.
By Lucas' Theorem, we get
$$
\begin{array}{l}
\mathrm{C}_{n}^{i_{1}} \equiv \mathrm{C}_{n_{0}}^{n_{0}}(\bmod p), \mathrm{C}_{n}^{i_{2}} \equiv \mathrm{C}_{n_{1}}^{1} \mathrm{C}_{n_{0}}^{n_{0}}(\bmod p) \\
\mathrm{C}_{n}^{i_{3}} \equiv \mathrm{C}_{n_{1}}^{2} \mathrm{C}_{n_{0}}^{n_{0}}(\bmod p)
\end{array}
$$

Thus, $1 \equiv n_{1} \equiv \frac{n_{1}\left(n_{1}-1\right)}{2}(\bmod p)$.
Clearly, this is not true.
Therefore, $p=2$.
Lemmas 1 and 2 are proved.
Notice that the parity of $\mathrm{C}_{n}^{i}+f(i)$ is fixed, i.e.,
$$
\begin{array}{l}
\mathrm{C}_{n}^{i}+f(i) \equiv \mathrm{C}_{n}^{j}+f(j)(\bmod 2)(0 \leqslant i < j \leqslant n).
\end{array}
$$

From $n > p^2$, we know $5 \leqslant n < 2020$, a total of 8 choices.
In summary, there are 8 pairs of $(n, p)$ that meet the requirements.